Let $(X_n,d_n),(Y,d_Y)$ be compact metric spaces for $n\in\mathbb N$, $\pi_n\in C(X_n,Y)$, $$\iota_nf:=f\circ\pi_n\;\;\;\text{for }f\in C(Y)$$ for $n\in\mathbb N$, $f_n\in C(X_n)$ for $n\in\mathbb N$ and $f\in C(Y)$.
Assume $$|f_n(x_n)-f(y)|\xrightarrow{n\to\infty}0\tag1$$ for all $x_n\in X_n$ for $n\in\mathbb N$ and $y\in Y$ with $\pi_n(x_n)\xrightarrow{n\to\infty}y$. Are we able to conclude $$\left\|f_n-\iota_nf\right\|_\infty\xrightarrow{n\to\infty}0\tag2?$$
Note that the result holds if $E_n=E$ and $\pi_n$ is the identity for all $n\in\mathbb N$.
EDIT: Maybe we are able to use the known result from the link. Note that the condition in $(1)$ is equivalent to $$|(f_n-\iota_nf)(x_n)|\xrightarrow{n\to\infty}0\tag3$$ for all $x_n\in X_n$ such that $(\pi_n(x_n))_{n\in\mathbb N}$ is convergent.
Note that $$X:=\prod_{n\in\mathbb N}X_n$$ equipped with $$d_X(x,x'):=\sum_{n\in\mathbb N}2^{-n}d_n(x_n,x'_n)\;\;\;\text{for }x,x'\in X$$ is a compact metric space. By the result from the link, if $(g_n)_{n\in\mathbb N}\subseteq C(X)$ and $g\in C(X)$, then $$\left|g_n\left(x^{(n)}\right)-g\left(x^{(n)}\right)\right|\xrightarrow{n\to\infty}0\tag4$$ for all convergent $\left(x^{(n)}\right)_{n\in\mathbb N}\subseteq X$ (or, equivalently, $$\left|g_n\left(x^{(n)}\right)-g(x)\right|\xrightarrow{n\to\infty}0\tag5$$ for all $\left(x^{(n)}\right)_{n\in\mathbb N}\subseteq X$ and $x\in X$ with $d_X\left(x^{(n)},x\right)\xrightarrow{n\to\infty}0$), then $$\left\|g_n-g\right\|_\infty\xrightarrow{n\to\infty}0\tag6.$$
Now, let $$g_n(x):=(f_n-\iota_nf)(x_n)\;\;\;\text{for }x\in X$$ for $n\in\mathbb N$. We would need to show that if $\left(x^{(n)}\right)_{n\in\mathbb N}\subseteq X$ is convergent, then $$\left|g_n\left(x^{(n)}\right)\right|\xrightarrow{n\to\infty}0\tag7.$$ In order to do so, we would need to show that $\left(\pi_n\left(x^{(n)}_n\right)\right)_{n\in\mathbb N}$ is convergent and proceed accordingly. Are we able to do so?