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
Hint : Show that $\liminf f_n$ is measurable. The proof for $\limsup f_n$ is similar, and $$ \{ x \in X \, | \, \lim_{n \to \infty} f_n \text{ exists} \} = \{ x \in X \, | \, \liminf_{n\to \infty} f_n = \limsup_{n\to \infty} f_n\}. $$ Hope that helps,