Yes, it's rather trivial. You might note the following fact: if $f$ is measurable, $\mu$ is complete, and $f = g$ $\mu$-a.e., then $g$ is measurable. This is valid for functions from any complete measure space to any other measurable space. (Proving it is a good exercise.) Essentially, when you have a complete measure, you can change a function arbitrarily on a null set without breaking its measurability.
Now suppose $f_n \to f$ $\mu$-a.e., with each $f_n$ measurable. Let $A$ be the set of $x$ such that $f_n(x) \to f(x)$. Note $A$ is a measurable set since $A^C$ is $\mu$-null and hence measurable (by completeness of $\mu$). Let $$g_n = \begin{cases}f_n(x), & x \in A \\ 0, & x \notin A\end{cases}$$ Then clearly $g_n$ is measurable. Now $g_n(x)$ converges for all $x$, so the limit $g(x) = \lim_{n \to \infty} g_n(x)$ is measurable, as you know. But $g(x) = f(x)$ on $A$, so $f = g$ a.e. By the fact above, $f$ is measurable.
If your underlying measure space is not complete, certainly things can go wrong. For a trivial example, let $B$ be any set which is $\mu$-null but not measurable. If $f_n$ is the zero function for all $n$, we have $f_n \to 1_B$ a.e., but $1_B$ is not a measurable function.
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 $C$.
Best Answer
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 $X$).