Let $X$ be a metric space and let $f_{n}:X\to\mathbb{R}$ be a uniformly convergent sequence converging to a continuous function $f:X\to\mathbb{R}$ on $X$ . Let $\left\{ x_{n}\right\} _{n=1}^{\infty}\subseteq X$ be a sequence converging to $x^{*}\in X$ , show that $f_{n}\left(x_{n}\right)$ converges to $f\left(x^{*}\right)$ .
Proof: Given $\varepsilon>0$ by continuity of $f$ there exists a $\delta>0$ such that $$d\left(x,x^{*}\right)<\delta\Longrightarrow\left|f\left(x\right)-f\left(x^{*}\right)\right|<\frac{\varepsilon}{2}$$ From convergence of $\left\{ x_{n}\right\} _{n=1}^{\infty}$ to $x^{*}$ there exists an $N_{1}\in\mathbb{N}$ such that for all $n\geq N_{1}$ we have $d\left(x_{n},x^{*}\right)<\delta$ and thus by continuity $\left|f\left(x_{n}\right)-f\left(x^{*}\right)\right|<\frac{\varepsilon}{2}$ . From uniform convergence of $f_{n}$ there is an $N_{2}\in\mathbb{N}$ such that for all $n\geq N_{2}$ $\left|f_{n}\left(x_{n}\right)-f\left(x_{n}\right)\right|<\frac{\varepsilon}{2}$ and thus for all $n\geq\max\left\{ N_{1},N_{2}\right\}$ $$\left|f_{n}\left(x_{n}\right)-f\left(x^{*}\right)\right|\leq\left|f_{n}\left(x_{n}\right)-f\left(x_{n}\right)\right|+\left|f\left(x_{n}\right)-f\left(x^{*}\right)\right|<\varepsilon$$ Proving the claim.
Now I wish to show this is also true if the convergence is only locally uniform. However, I run into the problem that I no longer have an $N_{2}$ such that for all $n\geq N_{2}$ $\left|f_{n}\left(x_{n}\right)-f\left(x_{n}\right)\right|<\frac{\varepsilon}{2}$ . At most for all $n$ I know there is neighborhood of $x_{n}$ where the $f_{n}$ converges uniformly and thus I can select an $N_{x_{n}}$ such that for all $n\geq N_{x_{n}}$ I have $\left|f_{n}\left(x_{n}\right)-f\left(x_{n}\right)\right|<\frac{\varepsilon}{2}$ . But now I can't take the maximum of all such $N_{x_{n}}$ to achieve my goal. Compactness of $X$ would solve this easily but that is not part of the claim I'm trying to prove. I'd appreciate some help.
You can reduce this problem to the case of uniform convergence. Choose $r>0$ sufficiently small so that $f_n$ converges uniformly on $B_r(x^*)$. Now note that all but finitely many $x_n$ lie in $B_r(x^*)$.
If you want to be more formal on the last part: Choose $N$ sufficiently large so that $x_n \in B_r(x^*)$ for all $n > N$. Now define the new sequencse $y_n = x_{N + n}$ and $g_n = f_{n + N}\mid_{B_r(x^*)}$. Then $g_n$ converges uniformly to $f$, $y_n$ converges uniformly to $x^*$ and by the first part $f_{n+N}(x_{n+N}) = g_n(y_n) \to f(x^*)$.