Does $\lim_{n \to \infty} \sup_{x \in X} f_n(x) = \sup_{x \in X} \lim_{n \to \infty} f_n(x)$?

No, in general you cannot do that. Imagine if $X$ is $\mathbb{R}$ and $f_n(x)$ is zero except on $[n,n+1]$ where it is $1$. Work out both sides of the equation in this case...

No. Consider, for example $$f_n(x) = \frac{1}{(x-n)^2+1}$$ and $X=\mathbb R$. The supremum of each $f_n$ (and thus the limit of the suprema) is $1$, but the pointwise limit at each $x$ (and thus the supremum of the limits) is $0$.

You'll have better luck if you can assume that the $f_n$s converge uniformly on $X$.

$$f_n(x)\leq \sup_{x\in A} f_n(x) $$ $$\Rightarrow {\liminf}_{n\rightarrow \infty} f_n(x)\leq {\liminf}_{n\rightarrow \infty}\sup_{x\in A}f_n(x) $$ $$\Rightarrow \sup_{x\in A}{\liminf}_{n\rightarrow \infty} f_n(x)\leq {\liminf}_{n\rightarrow \infty}\sup_{x\in A}f_n(x) $$