The claim is false. We will construct a counterexample with $X = Y = (C[0,1], \| \cdot \|_{\infty})$ and $|\cdot | = \| \cdot \|_{L_{1}}$. This is the original counterexample constructed for this question and it is inspired by a generalisation of a hint to Exercise 6.13 in Functional Analysis, Sobolev Spaces and Partial Differential Equations by Brezis. To outline the initial construction, we take a countable dense sequence $(t_{k})_{k\in\mathbb{N}}$ in $[0,1]$ and a sequence $(g_{k})_{k\in\mathbb{N}}$ of suitable functions with (almost) disjoint supports. We then define a bounded linear operator $T$ on $C[0,1]$ given by
\begin{equation}
Tf = \sum_{k=1}^{\infty} 2^{1-k} f(t_{k}) g_{k} . \tag{$\ast$}
\end{equation}
It is shown the linear operator $T$ is injective and compact, while it is also shown there is no $C > 0$ such that
\begin{equation}
\|Tf\|_{\infty} \leq \tfrac{1}{2} \|f\|_{\infty} + C \|f\|_{L_{1}}
\end{equation}
for all $f\in C[0,1]$. After that, it is briefly mentioned how we can take $Y$ as a sequence space instead of $Y = C[0,1]$ and modify the counterexample to that setting. In particular, the case where $Y = c_{0}$ has a proof that is perhaps a bit simpler than the original argument.
Let $(t_{k})_{k\in\mathbb{N}}$ be a sequence that is dense in $[0,1]$. Also let $(g_{k})_{k\in\mathbb{N}}$ be a sequence in $C[0,1]$ such that $\|g_{k}\|_{\infty} = 1$ for all $k\in\mathbb{N}$ and
\begin{equation}
\{ t\in [0,1] : g_{j}(t) \neq 0 \} \cap \{ t\in [0,1] : g_{k}(t) \neq 0 \} = \emptyset \tag{1}
\end{equation}
for all $j,k\in \mathbb{N}$ with $j \neq k$. An example of such a sequence is $(g_{k})_{k\in\mathbb{N}}$, where $g_{k} \colon [0,1] \to \mathbb{R}$ is defined by $g_{k}(t) := \max\{1 - 2^{k+1}|t - \tfrac{3}{2^{k+1}}| , 0\}$ for each $k\in\mathbb{N}$. There is a helpful comment to this answer by Just a user that points out some of the geometric properties of these $g_{k}$ functions that may help the reader follow along with the proof. See here for the comment.
Define $T\colon C[0,1] \to C[0,1]$ by
\begin{equation}
Tf := \sum_{k=1}^{\infty} 2^{1-k} f(t_{k}) g_{k}.
\end{equation}
It is straightforward to check that $T$ is a bounded linear operator on $(C[0,1], \| \cdot \|_{\infty})$. It can in fact be shown directly using $(1)$ that $\|T\| = 1$, but we will not need to make use of this. We will show below that $T$ is injective and compact.
We first show that $T$ is injective. Let $f\in C[0,1]$ such that $Tf = 0$. Let $k\in \mathbb{N}$. Since $\|g_{k}\|_{\infty} = 1 > 0$ there is some $t\in [0,1]$ such that $g_{k}(t) \neq 0$. By $(1)$ we have $g_{j}(t) = 0$ for all $j\in\mathbb{N}\setminus\{k\}$. It follows that
\begin{equation}
2^{1-k} f(t_{k}) g_{k}(t) = (Tf)(t) = 0.
\end{equation}
Since $2^{1-k} g_{k}(t) \neq 0$ we have $f(t_{k}) = 0$. As $k\in\mathbb{N}$ was arbitrary we see that $f(t_{k}) = 0$ for all $k\in\mathbb{N}$. But since $\{t_{k} : k\in\mathbb{N}\}$ is dense in $[0,1]$, it follows from the continuity of $f$ that $f(t) = 0$ for all $t\in [0,1]$. Hence $f = 0$. This shows that $T$ is injective.
We now show $T$ is compact. Let $n\in\mathbb{N}$. Define $T_{n} \colon C[0,1] \to C[0,1]$ by
\begin{equation}
T_{n}f := \sum_{k=1}^{n} 2^{1-k} f(t_{k}) g_{k}.
\end{equation}
Then $T_{n}$ is a finite rank linear operator on $C[0,1]$. Let $f\in C[0,1]$. Using $(1)$, we have
\begin{equation}
|T_{n}f - Tf| ={} \Big| \sum_{k=n+1}^{\infty} 2^{1-k} f(t_{k}) g_{k} \Big| \leq \sup \{2^{1-k} |f(t_{k})| : k\in \mathbb{N} \text{ and } k > n \} \leq 2^{-n} \|f\|_{\infty} . \tag{2}
\end{equation}
It follows from $(2)$ that $\|T_{n} - T\| \leq 2^{-n}$. This shows that $T$ is in the norm closure of the set of finite rank linear operators and is consequently compact.
We now show there does not exist $C > 0$ such that
\begin{equation}
\| Tf \|_{\infty} \leq \tfrac{1}{2} \|f\|_{\infty} + C \|f\|_{L_{1}} \tag{3}
\end{equation}
for all $f\in C[0,1]$, where $\|f\|_{L_{1}} = \int_{0}^{1} |f|$ for each $f\in C[0,1]$. Since $\|f\|_{L_{1}} \leq \|f\|_{\infty}$ for every $f\in C[0,1]$, showing that $(3)$ fails for every $C > 0$ would provide the desired counterexample. Let $C > 0$. Since $\|g_{0}\|_{\infty} = 1$ there is $s_{0}\in [0,1]$ such that $|g_{0}(s_{0})| = 1$. Take $f\in C[0,1]$ such that $\|f\|_{\infty} = 1$, $f(s_{0}) = 1$ and $\int_{0}^{1} |f| < \tfrac{1}{2C}$. An example of such a function is $f\colon [0,1] \to \mathbb{R}$ defined by $f(t) := \max\{1 - 4C|t - s_{0}|, 0\}$. We have
\begin{equation}
\tfrac{1}{2} \|f\|_{\infty} + C \|f\|_{L_{1}} < \tfrac{1}{2} + C \cdot \tfrac{1}{2C} = 1 \tag{4}
\end{equation}
while
\begin{equation}
|(Tf)(s_{0})| = |f(s_{0}) g_{0}(s_{0})| = 1 .
\end{equation}
Consequently, it follows that $\|Tf\|_{\infty} \geq 1$ and hence, using $(4)$, that
\begin{equation}
\|Tf\|_{\infty} \geq 1 > \tfrac{1}{2} \|f\|_{\infty} + C \|f\|_{L_{1}} .
\end{equation}
As $C > 0$ was arbitrary, we have a counterexample.
Here are some additional comments. Upon further inspection, it came to my attention that instead of taking the codomain of $T$ as $C[0,1]$ as done in $(\ast )$, we could have instead modified the definition of $T$ and instead mapped into a sequence space. Perhaps the simplest example of this is the following. Take a dense sequence $(t_{k})_{k\in\mathbb{N}}$ in $[0,1]$ and define $S\colon C[0,1] \to c_{0}$ by
\begin{equation}
Sf := (2^{1-k} f(t_{k}))_{k\in\mathbb{N}} .
\end{equation}
The advantage here is that we no longer need to find a suitable sequence of functions $(g_{k})_{k\in\mathbb{N}}$ as in the definition of $(\ast )$. As in the above proof, the map $S$ is an injective compact operator and there is no $C > 0$ such that
\begin{equation}
\|Sf\|_{c_{0}} \leq \tfrac{1}{2} \|f\|_{\infty} + C \|f\|_{L_{1}}
\end{equation}
for all $f\in C[0,1]$. We can also take $\ell_{\infty}$ in place of $c_{0}$ in the above definition of $S$ and all the details are exactly the same as in the case with $c_{0}$.
Another important example of this generalisation is to take $p\in [1, \infty )$ as well as a dense sequence $(t_{k})_{k\in\mathbb{N}}$ in $[0,1]$ and define $S\colon C[0,1] \to \ell_{p}$ by
\begin{equation}
Sf := \big( 2^{-\tfrac{k}{p} } f(t_{k}) \big)_{k\in\mathbb{N}} .
\end{equation}
As done in the other cases, it is shown similarly that $S$ is an injective compact operator. This time, however, it is simpler to show that there is no $C > 0$ such that
\begin{equation}
\|Sf\|_{\ell_{p}} \leq \tfrac{1}{4} \|f\|_{\infty} + C \|f\|_{L_{1}} \tag{5}
\end{equation}
for all $f\in C[0,1]$. We show $(5)$ in detail. Let $C > 0$. Take $f\in C[0,1]$ such that $f(t_{1}) = 1$, $\|f\|_{\infty} = 1$ and $\int_{0}^{1} |f| < \tfrac{1}{4C}$. We have
\begin{equation}
\tfrac{1}{4} \|f\|_{\infty} + C \|f\|_{L_{1}} < \tfrac{1}{4} + C \cdot \tfrac{1}{4C} = \tfrac{1}{2} \tag{6}
\end{equation}
and
\begin{equation}
\| Sf \|_{\ell_{p}}^{p} = \sum_{k=1}^{\infty} |2^{-\tfrac{k}{p} } f(t_{k}) |^{p} \geq |2^{-\tfrac{1}{p} } f(t_{1})|^{p} = 2^{-1} \geq (2^{-1})^{p} , \tag{7}
\end{equation}
so it follows by combining $(6)$ and $(7)$ that
\begin{equation}
\| Sf \|_{\ell_{p}} \geq \tfrac{1}{2} > \tfrac{1}{4} \|f\|_{\infty} + C \|f\|_{L_{1}} .
\end{equation}
This shows there is no $C > 0$ such that $(5)$ holds for all $f\in C[0,1]$. Note that by taking $p\in (1, \infty )$ we can find a counterexample where the space $Y$ in the original question is reflexive, and by taking $p = 2$ we can even take the space $Y$ in the original question as a Hilbert space.
Finally, the reader is encouraged to take a look at this related result. Note that the linked result is not applicable to this question because we cannot typically find a suitable Banach space $(Y, \|\cdot\|_{Y})$ to apply the result.
