A proof of a positive answer for the question was given to me by professor Antonio Avilés.

Let's take $D \subset Y$ a countable dense subset of $Y$ and for every $y \in D$ and $m \in \Bbb N$ let's take some $\gamma_{y, m} \in \Gamma$ such that
$$\| y \|_{\Gamma} - \dfrac{1}{m} \leq |y (\gamma_{y,m})| \leq \| y \|_{\Gamma} \quad (1)$$
where I'm denoting by $\| \|_{\Gamma}$ the sup norm in $\ell_\infty(\Gamma)$.

Then, if we denote by $Q = \{\gamma_{y,m}: y \in D, m \in \Bbb N\}$ we have that $\ell_\infty (Q)$ is isometric to $\ell_\infty$ and that the map $u: Y \rightarrow \ell_\infty (Q)$ given by
$$u(x)(\delta) = x(\delta), \forall \delta \in Q$$
for every $x \in Y$ is a linear isometric embedding.

The linearity is clear and it is also clear that
$$\|u(x)\|_Q = \sup_{\delta \in Q} |u(x)(\delta)| = \sup_{\delta \in Q} |x(\delta)| \leq \|x\|_\Gamma \quad (2)$$
For the other inequeality, let $x \in Y$ fixed and take any $\varepsilon > 0$. Then, there exists some natural number $m$ such that $1/m < \varepsilon /2$ and some $y \in D$ such that
$$\| x - y \|_\Gamma < \dfrac{\varepsilon}{4} \quad (3)$$
Then we have that
$$|y(\gamma_{y,m})| \le |y(\gamma_{y,m}) - x(\gamma_{y,m})| + |x(\gamma_{y,m})| \leq \| x-y\|_\Gamma + |x(\gamma_{y,m})| \quad (4)$$
and so by (1), (3) and (4)
\begin{align}
\|x\|_\Gamma & \leq \|x-y\|_\Gamma + \|y\|_\Gamma < \|x-y\|_\Gamma + |y(\gamma_{y,m})| + \dfrac{1}{m} \leq |x(\gamma_{y,m})| + 2 \|x-y\|_\Gamma + \dfrac{\varepsilon}{2} < \\
& < \sup_{\delta \in Q} |x(\delta)| + \varepsilon = \|u(x)\|_Q + \varepsilon
\end{align}

As the $\varepsilon > 0$ is arbitrary we obtain that
$$\|x\|_\Gamma \leq \|u(x)\|_Q \quad (5)$$
By (2) and (5) it is proven that $u$ is a linear isometric embedding.

Let now $\pi_\gamma: \ell_\infty(\Gamma) \rightarrow \Bbb R$ be the projection over $\gamma \in \Gamma$, that is,
$$\pi_\gamma(x) = x(\gamma), \forall x \in \ell_\infty(\Gamma)$$
Then, if we denote by $\tilde Y = u (Y) \subset \ell_\infty (Q)$, we can define for every $\gamma \in \Gamma$ the maps $\pi_\gamma \circ u^{-1}: \tilde Y \rightarrow \Bbb R$ and extend them to $\tilde \pi_\gamma: \ell_\infty(Q) \rightarrow \Bbb R$ in such a way that:

- $\|\tilde \pi_\gamma\| \leq 1$
- If $\gamma \in Q$, then $\tilde \pi_\gamma$ is the projection over $\gamma$ on $\ell_\infty(Q)$, that is, $\tilde \pi_\gamma(z) = z(\gamma),\, \forall z \in \ell_\infty(Q)$

We can do that because if $\gamma \not \in Q$ then the extension is guaranteed by the Hahn-Banach theorem (taking into account that $\|\pi_\gamma \circ u^{-1}\| \leq \|\pi_\gamma\| \|u^{-1}\| = \|\pi_\gamma\| \leq 1$). And, if $\gamma \in Q$, then it is easy to see that the given projection have norm less than one and it is in fact an extension of $\pi_\gamma \circ u^{-1}$ because if $z \in \tilde Y$, then there exists some $x \in Y$ such that $z=u(x)$ and so
$$\tilde \pi_\gamma(z) = z(\gamma) = u(x)(\gamma) = x(\gamma) = (\pi_\gamma \circ u^{-1}) (u(x)) = (\pi_\gamma \circ u^{-1}) (z)$$
Lastly, let's define $w: \ell_\infty(Q) \rightarrow \ell_\infty(\Gamma)$ by
$$w(z)(\gamma) = \tilde \pi_\gamma(z),\, \forall z \in \ell_\infty(Q), \gamma \in \Gamma$$
We have that $w$ is clearly linear and satisfies that given $z \in \ell_\infty(Q)$:

- $\|w(z)\|_\Gamma = \sup_{\gamma \in \Gamma} |\tilde \pi_\gamma(z)| \leq \sup_{\gamma \in \Gamma} \|\tilde \pi_\gamma\| \|z\|_Q \leq \|z\|_Q$
- $\|z\|_Q = \sup_{\delta \in Q} |z(\delta)| = \sup_{\delta \in Q} |z(\delta)| = \sup_{\delta \in Q} |\tilde \pi_\delta(z)| \leq \sup_{\gamma \in \Gamma} |\tilde \pi_\gamma(z)| = \|w(z)\|_\Gamma$

So $w$ is also a linear isometric embedding.

Now, is we define $A = w(\ell_\infty(Q))$, as $\ell_\infty(Q)$ is linearly isometric to $\ell_\infty$ and $w$ is a linear isometry, we have that $A$ is linearly isometric to $\ell_\infty$ and
$$w(\tilde Y) \subset A \subset \ell_\infty(\Gamma) \quad (6)$$
But,
$$w(\tilde Y) = Y \quad (7)$$
because for every $x \in Y$ we have that
$$w(u(x))(\gamma) = \tilde \pi_\gamma (u(x)) = (\pi_\gamma \circ u^{-1}) (u(x)) = \pi_\gamma (x) = x(\gamma),\, \forall \gamma \in \Gamma$$
So, by (6) and (7) we conclude that $Y \subset A \subset \ell_\infty (\Gamma)$ as wanted.