The answer to your question is Yes provided the topology of $Y$ is such that for each non-empty open set $O\subset Y$ there is a strictly increasing sequence $(O_k)$ of open sets: $$ \overline O_k\subset O_{k+1}\subset O,\quad k=1,2,\ldots, $$ with $\cup_kO_k=O$. For suppose $(f_n)$ is a sequence of Borel functions from $X$ to $Y$ with pointwise limit $f$. With $O$ and the $O_k$ as above, $$ f^{-1}(O)=\cup_{k=1}^\infty\cup_{n=1}^\infty\cap_{m=n}^\infty f_m^{-1}(O_k). $$ This shows that $f^{-1}(O)$ is a measurable subset of $X$. Because the Borel $\sigma$-field on $Y$ is generated by the open sets, it follows that $f$ is Borel measurable.

In particular, the condition above is true if $Y$ is metrizable.

The answer to your question is Yes provided the topology of $Y$ is such that for each non-empty open set $O\subset Y$ there is a strictly increasing sequence $(O_k)$ of open sets: $$ \overline O_k\subset O_{k+1}\subset O,\quad k=1,2,\ldots, $$ with $\bigcup_kO_k=O$. For suppose $(f_n)$ is a sequence of Borel functions from $X$ to $Y$ with pointwise limit $f$. With $O$ and the $O_k$ as above, $$ f^{-1}(O)=\bigcup_{k=1}^\infty\bigcup_{n=1}^\infty\bigcap_{m=n}^\infty f_m^{-1}(O_k). $$ This shows that $f^{-1}(O)$ is a measurable subset of $X$. Because the Borel $\sigma$-field on $Y$ is generated by the open sets, it follows that $f$ is Borel measurable.

In particular, the condition above is true if $Y$ is metrizable.