Solution
= Solution
This is always true. Uniform convergence makes $(f_n)$ a <uniformly Cauchy sequence>: for every $\varepsilon>0$, there is $N$ such that
$$
d_Y(f_n(x),f_m(x))<\varepsilon
$$
for every $x\in A$ and $m,n\geq N$. Letting $x\to x_0$ through points of $A\setminus\{x_0\}$ and using the assumed limits gives
$$
d_Y(y_n,y_m)\leq\varepsilon.
$$
\b[Thus $(y_n)$ is a <Cauchy sequence>. Since $Y$ is a <complete metric space>, $(y_n)$ converges in $Y$.]