$\lim_{n \to \infty}f_n(x_n)=f(x)$ if $f_n \to f$ and $x_n \to x$?

This is only true if $f_n(x)$ converges uniformly to $f(x)$. In fact, it will break for any situation where $f_n(x)\rightarrow f(x)$ pointwise but not uniform on an interval $[L,R]$. In other words, there exists $\epsilon>0$ s.t. for all $N>0$, there exists a point $a$ such that $|f_n(a)-f(a)|>\epsilon$. Pick a sequence of $\epsilon_n$ and increasing $N_n$. This gives a sequence of $a_n$. By Bolzano-Weierstrass, pick a convergent subsequence of $a_n$ that converges, and call it $x_n$. Now follow through the proof.

Example: $f_n(x)=2^nx$ when $0\leq x\leq 2^{-n}$, $f_n(x)=2-2^nx$ when $2^{-n}\leq x\leq 2^{-n+1}$ and 0 elsewhere. Then $f_n(x)\rightarrow 0$. Yet pick $x_n=1/2^n$. Then $x_n\rightarrow 0$ but $f_n(x_n)=1$. Essentially you have a bump which is clamped to 0 on either side of $[0,1]$, it's squishing to the left but there's always a sequences which rides on the value 1.