Point-wise limit of finite valued functions

I think it is true (even in the more general setting of second countable topological spaces). Let $\{A_n\}_{n\in\mathbb{N}}$ be a countable basis of the topology of $X$. For ${n\in\mathbb{N}}$ let $\Pi_n$ be the finite partition of $X$ generated by the sets $A_1,\dots, A_n$, and let $f_n:X\to X$ be constant on each $E\in\Pi_n$, with $f_n(E)\in E$. Then $f_n(x)\to x$ for any $x\in X$. (Indeed, for any $x\in X$ and for any nbd $U$ of $x$ there is $n_0\in\mathbb{N}$ such that $x\in A_{n_0}\subset U$. Then by construction, for any $n\ge n_0$ there holds $f_n(x)\in U$).

To answer a question raised in comments: if $X$ is merely separable then this can fail. Let $X = \mathbb{R}^\mathbb{R}$ with the product topology, which is a Hausdorff topological vector space. Then $X$ is separable. (Any product of continuum many separable spaces is separable. Alternatively, viewing $X$ as the space of all functions from $\mathbb{R}$ to $\mathbb{R}$, one can show that the set of polynomials with rational coefficients is dense.)

Now let $f_n : X \to X$ be a sequence of finite-range functions converging pointwise to some $f$. Let $A = \bigcup_n f_n(X)$ be the union of the images of the $f_n$, so that $A$ is countable. Then the number of sequences of elements of $A$ has cardinality at most $(\aleph_0)^{\aleph_0} = \mathfrak{c}$, so that the set $B$ of possible limits of sequences from $A$ likewise has cardinality at most $\mathfrak{c}$. But the image of $f$ is contained in $B$. Since $X$ has cardinality $\mathfrak{c}^\mathfrak{c} > \mathfrak{c}$, the function $f$ cannot be surjective.

I'll assume that $X$ is Hausdorff.

A Hausdorff topological vector space is metrizable iff it is first countable. Hence, $X$ is metrizable and has a dense subset $\{x_n\}_{n=1}^{\infty}$. Define $f_n(x)$ as an arbitrary closest of $x_1,...,x_n$ to $x$. Then $f_n(x)\to x$. Indeed, $d(x,f_n(x))$ does not increase, and for every $\varepsilon>0$ there is $n$ such that $d(x,f_n(x))\le d(x,x_n)<\varepsilon$.