I have to prove that a projection operator maps $L^\infty_\alpha$ onto $F^\infty_\alpha$. Before stating the problem more precisely, I define the objects at stake: given $\alpha>0$, we set $d\lambda_\alpha(z):=\frac{\alpha}{\pi}e^{-\alpha|z|^2}dA(z)$, a (probability) measure on $\mathbb{C}$, where $dA(z)$ is the Lebesgue measure on $\mathbb{C}$, i.e. $dA(z)=dxdy$. We set for $p\in[1,+\infty)$
\begin{equation} L^p_\alpha:=\Big\{f:\mathbb{C}\rightarrow\mathbb{C} \ : \ f \ is \ measurable \ and \ \Vert f\Vert_{p,\alpha}:=\Big(\frac{p\alpha}{2\pi}\int_{\mathbb{C}}|f(z)|^pe^{-p\alpha|z|^2/2}dA(z)\Big)^{1/p}<\infty\Big\}=L^p(\mathbb{C},d\lambda_{p\alpha/2}), \end{equation}
\begin{equation} L^\infty_\alpha:=\Big\{f:\mathbb{C}\rightarrow\mathbb{C} \ : \ f \ is \ measurable \ and \ \Vert f\Vert_{\infty,\alpha}:=\sup_{z\in\mathbb{C}}|f(z)|e^{-\alpha|z|^2/2}<\infty\Big\} \end{equation}
and we set $F^p_\alpha:=Hol(\mathbb{C})\cap L^p_\alpha$. It is easy to see that $F^p_\alpha\hookrightarrow F^\infty_\alpha$ for all $p\in[1,+\infty)$.
$F^2_\alpha$ is a reproducing kernel Hilbert space with respect to the inner product \begin{equation} \langle f,g\rangle:=\int_\mathbb{C}f(z)\overline{g(z)}d\lambda_\alpha(z) \end{equation} with kernel $K_\alpha(z,w)=e^{\alpha w\overline{z}}$.
One can check that functions in the form $\phi_{\gamma,w_1,\ldots,w_N}(z)=\sum_{j=1}^Na_je^{\gamma z\overline{w}_k}$, for $N\in\mathbb{N}$, $w_1,\ldots,w_N,a_1,\ldots,a_N\in\mathbb{C}$ and $\gamma>0$, is dense in $F^p_\alpha$ for all $\alpha>0$ and $p\in[1,+\infty)$.
For all $\alpha>0$, we define $P_\alpha$ as the integral operator \begin{equation} P_\alpha f(z):=\int_\mathbb{C}e^{\alpha z\overline{w}}f(z)d\lambda_\alpha(w). \end{equation} As it is easy to see, $P_\alpha$ is the projection of $L^2(\mathbb{C},d\lambda_\alpha)=L^2_\alpha$ onto $F^2_\alpha$.
$P_\alpha:L^p_\alpha\rightarrow F^p_\alpha$ is a bounded surjective projection for all $\alpha>0$ and for all $p\in[1,+\infty]$.
Using the density property above it is easy to see that $P_\alpha|_{F^p_\alpha}$ is the identity for all $\alpha>0$ and $1\leq p<\infty$. Moreover, $P_\alpha$ is bounded from $L^p_\alpha$ to $F^p_\alpha$ for all $p$.
Using the kernel property and Fubini's theorem, it is easy to see that $P^2_\alpha=P_\alpha$ on $F^\infty_\alpha$ ($\Longrightarrow$ $P^2_\alpha=P_\alpha$ on all $F^p_\alpha$).
I have to prove that $P_\alpha:L^\infty_\alpha\rightarrow F^\infty_\alpha$ and it is onto. I proved that $P_\alpha(L^\infty_\alpha)\subseteq L^\infty_\alpha$, so it remains to prove that $P_\alpha f\in Hol(\mathbb{C})$ for all $f\in F^\infty_\alpha$ and the surjectivity.
For the holomorphy, I thought to differentiate under the integral, since $f\in L^\infty_\alpha$ implies that \begin{equation} \Big|\frac{d}{dz}\Big(e^{\alpha\overline{z}w}f(w)e^{-\alpha|w|^2}\Big)\Big|\leq C|w|e^{-\alpha|w|^2/2}\in L^1(\mathbb{C},dA(z)). \end{equation} I think this' enough to prove the holomorphy. Otherwise, one could use Morera's theorem to reach the same conclusion.
However, the surjectivity, which is clear in the $p\neq\infty$ case due to the density argument, is not that easy to prove in this situation. My idea was that of proving that $P_\alpha|_{F^\infty_\alpha}$ is the identity on $F^\infty_\alpha$ as well. This is equivalent to claiming that $e^{\alpha z\overline{w}}$ acts as a kernel also on $F^\infty_\alpha$ as, in this case, it would be \begin{equation} P_\alpha f(z)=\int_\mathbb{C}f(w)e^{\alpha z\overline{w}}d\lambda_\alpha(w)=\langle f, e^{\pi\overline{z}\cdot}\rangle=f(z). \end{equation} But I don't know how to prove this formula for $F^\infty_\alpha$. Duality arguments are precluded here.
The whole thing is the last theorem in K. Zhu's work "Analysis on Fock spaces", Chapter 1.2, however the author reduces the proof to that of $P\alpha:L^p_\alpha\rightarrow F^p_\alpha$ continuously and does not prove the other conditions. Maybe, there is some property I missed that would make the argument trivial.
The kernel formula holds for all $f\in F^\infty_\alpha$ because of the fact that $\exists\beta>0$ s.t. $F^\infty_\alpha\hookrightarrow F^2_\beta$. In fact, for all $f\in F^\infty_\alpha$ \begin{equation} \Vert f\Vert_{2,\beta}^2=\frac{\beta}{\pi}\int_{\mathbb{C}}|f(z)|^2e^{-\beta|z|^2}dA(z)\leq\frac{\beta}{\pi}\Vert f\Vert_{\infty,\alpha}^2\int_{\mathbb{C}}e^{(\alpha-\beta)|z|^2}dA(z)\leq C\Vert f\Vert_{\infty,\alpha}^2 \end{equation} if $\beta>\alpha>0$. Fixed $\beta>\alpha$, using the reproducing kernel property of $F^2_\beta$ three times ((1)-(3)) and Fubini (F):
\begin{equation} \int_{\mathbb{C}}f(w)e^{\alpha\overline{w}z}d\lambda_\alpha(w)\underset{(1)}{=}\int_{\mathbb{C}}\int_{\mathbb{C}}f(\zeta)e^{\beta\overline\zeta w}d\lambda_\beta(\zeta)e^{\alpha\overline{w}z}d\lambda_\alpha(w)\underset{(F)}{=}\int_{\mathbb{C}}f(\zeta)\underbrace{\int_{\mathbb{C}}e^{\beta\overline\zeta w}e^{\alpha\overline{w}z}d\lambda_\alpha(w)}_\text{$\underset{(2)}{=}e^{\beta\overline{\zeta}z}$}d\lambda_\beta(\zeta)\underset{(3)}{=}f(z). \end{equation}
Since $F^p_\alpha\hookrightarrow F^\infty_\alpha$ for all $p\in(0,+\infty)$, the reproducing kernel formula holds on $F^p_\alpha$ for all $\alpha>0$ and $p\in(0,+\infty]$ and we are done.