To me it seems that there are two crucial factors for your proof. First is the space being regular, by which I mean just that any closed set and an outside point can be separated by disjoint neighbourhoods, without requiring $X$ to be $T_0$. Second, that any closed set has a countable neighbourhood base.
Given the above assumptions, we can use the same argument. If $C\subseteq X$ is closed, let $\{V_j\}_{j\in\mathbb{N}} \subseteq \mathcal{T}$ be its neighbourhood base. Now we prove that:
$$ f^{-1}(C)=\bigcap_{j\in\mathbb{N}}\bigcup_{k\in\mathbb{N}}\bigcap_{n\geq k}f_n^{-1}(V_j) $$
If $\omega\in f^{-1}(C)$, then $\{f_n(\omega)\}$ is eventually in any neighbourhood of $C$. Hence for all $j\in \mathbb{N}$, there exists $k_j \in \mathbb{N}$, such that $$\omega\in \bigcap_{n\geq k_j}f_n^{-1}(V_j)$$ implying that $\displaystyle{\omega\in \bigcup_{k\in\mathbb{N}}\bigcap_{n\geq k}f_n^{-1}(V_j)}$. $\;$ Hence we obtain:$\;$ $\displaystyle{\omega\in \bigcap_{j\in\mathbb{N}}\bigcup_{k\in\mathbb{N}}\bigcap_{n\geq k}f_n^{-1}(V_j)}$
Conversely, if $\displaystyle{\omega\in \bigcap_{j\in\mathbb{N}}\bigcup_{k\in\mathbb{N}}\bigcap_{n\geq k}f_n^{-1}(V_j)}$, then for each $j\in\mathbb{N}$: $\;\displaystyle{\omega\in \bigcap_{n\geq k_j}f_n^{-1}(V_j)}$ for some $k_j\in\mathbb{N}$, meaning that $\{f_n(\omega)\}$ is eventually in $V_j$. Suppose now that $f(\omega)\in C^c$ and let $W_1$ and $W_2$ be some neighbourhoods of $\omega$ and $C$ respectively. Since there exists some $s\in\mathbb{N}$ such that $C\subseteq V_s \subseteq W_2$ and since $f(\omega)$ is a limit of $\{f_n(\omega)\}$, it follows that eventually $\{f_n(\omega)\}$ is in both $W_1$ and $W_2$, meaning that $W_1 \cap W_2 \neq \varnothing$. Since the neighbourhoods are arbitrary, it means that $f(\omega)$ cannot be separated from $C$, contradicting regularity. Therefore $f(\omega)$ must be in $C$.
Interestingly, the above assumptions do not imply that $X$ is Hausdorff, unless it also happens to be $T_0$, in which case the countability condition will also be stronger than first countability.
EDIT (Weaker assumption)==========================================
Let $\mathcal{B}_X$ be the Borel sigma-algebra of a topological space $(X,\mathcal{T})$. In what follows $\varphi(\mathscr{C})$ denotes the filter generated by a subbase $\mathscr{C} \subset \mathcal{P}(X)$ and $\mathscr{N}(A)$ - the neighbourhood filter of a subset $A$
Assumptions:
- $\mathcal{T}$ is regular (not assuming $T_0$)
- For any nonempty closed $C \subseteq X$ there exists $\{V_j\}_{j\in\mathbb{N}} \subseteq \mathscr{N}(C) \cap \mathcal{B}_X$, such that any convergent filter containing $\{V_j : j\in\mathbb{N}\}$ contains $\mathscr{N}(C)$
Note that such $\{V_j : j\in\mathbb{N}\}$ is necessarily a filter subbase, since it has the finite intersection property, so there do exist filters that contain it. However it is not necessarily a base for $\mathscr{N}(C)$.
As before, since each $V_j$ is a neighbourhood of $C$, we have
$$f^{-1}(C) \subseteq \bigcap_{j\in\mathbb{N}} \bigcup_{k\in\mathbb{N}} \bigcap_{n\geq k} f^{-1}_n(V_j)$$
On the other hand
$$\omega \in \bigcap_{j\in\mathbb{N}} \bigcup_{k\in\mathbb{N}} \bigcap_{n\geq k} f^{-1}_n(V_j) \implies \{f_n(\omega)\} \text{ is eventually in each }V_j \implies$$
$$\implies \mathscr{F}_\omega =: \varphi \Big( \Big\{ \{f_n(\omega) : n\geq k\} : k\in\mathbb{N}\Big\} \Big) \supseteq \{V_j : j\in\mathbb{N}\} \implies$$
$$\implies \mathscr{F}_\omega \supseteq \mathscr{N}(C) \quad \text{ since } \mathscr{F}_\omega \text{ is convergent}$$
$$\implies \forall \quad U\in\mathscr{N}(C), W\in\mathscr{N}(f(\omega)): \Big( \{f_n\} \text{ is eventually in } U\cap W \implies U \cap W \neq \varnothing \Big)$$
$$\implies f(\omega) \in C \quad \text{by regularity assumption}$$
$$\\$$
The sets $\{C_{2^{-n}}\}$ in Shalop's proof satisfy the second assumption (which can be verified using continuity of $d(\cdot, C)$), while not necessarily being a neighbourhood base at $C$.
This is a really instructive question: apparently, completeness is the key!
First, let $X=Y=[0,1]$ with the usual topology. We define an example for a sequence of functions $f_n: X\rightarrow Y$, which converges everywhere point-wise. Then we ruin $Y$.
Let $f_1$ be constant $0$ on $[0,1/2[$ and constant $1/2$ on $[1/2,1]$.
Let $f_2$ be constant $0$ on $[0,1/4[$, constant $1/4$ on $[1/4,1/2[$, constant $1/2$ on $[1/2,3/4[$, and constant $3/4$ on $[3/4,1]$.
In general, divide $[0,1]$ into $2^n$ intervals of equal length (closed from the left and open from the right, except for the last one which is closed), and let $f_n$ be the step function whose value on each interval is the left endpoint.
It is clear that this sequence of functions converges point-wise to the identity function on $[0,1]$.
We keep these functions, but modify $Y$ now. Pick any non-measurable set $S\subseteq [0,1]$, and let $D$ consist of those rational numbers in $[0,1]$ whose denominator is a power of $2$.
Let $Y$ be the topological subspace of $[0,1]$ induced by $S\cup D$. This is of course a metric space (as it is a subspace of a metric space), but it is no longer complete!
Define the same sequence of functions as above; they are still measurable (and also functions $X\rightarrow Y$, because we were careful enough to put all elements of $D$ in $Y$).
However, the domain of convergence is $Y= S\cup D$: a sequence in $Y$ is convergent if and only if it is convergent as a sequence in $[0,1]$ with limit in $Y$.
As $S$ is not measurable and $D$ is countable, we have that $Y$ is not measurable (as a subset of $X$).
Best Answer
Thank you @OliverDÃaz so much for providing a reference containing a counter-example! Below is taken from Dudley's Real Analysis and Probability.
Now let $I=[0,1]$ with its usual topology. Then $I^I$ with product topology is a compact Hausdorff space by Tychonoff's Theorem (2.2.8). Such spaces have many good properties, but Theorem 4.2.2 does not extend to them (as range spaces), according to the following fact. Its proof assumes the axiom of choice (as usual, especially when dealing with a space such as $I^I$ ).
4.2.3. Proposition There exists a sequence of continuous (hence Borel measurable) functions $f_n$ from $I$ into $I^I$ such that for all $x$ in $I, f_n(x)$ converges in $I^I$ to $f(x) \in I^I$, but $f$ is not even Lebesgue measurable: there is an open set $W \subset I^I$ such that $f^{-1}(W)$ is not a Lebesgue measurable set in $I$.