Let $(X,d)$ be a metric space and let $f_j : X \to [-\infty , +\infty ]$ for all $j\in \mathbb{N}$.
Observe that since $X$ is metric, compactness is equivalent to sequential compactness.
Problem: If we fix an arbitrary compact $K\subset X$, I ask if it is always possible to have the existance of a sequence $(y_j) \subset K$ such that $$\liminf_{j\to +\infty} \inf_{x\in K} f_j(x) = \liminf_{j\to +\infty} f_j(y_j)$$
I need the compactness (sequential compactness) later on to find a subsequence of $(y_j)$ converging in $K$.
What I observed is that we don't have any kind of regularity of the $f_j$, so for example I could not conclude that $\inf_{x\in K}f_j(x) = f_j(x_{\textit{min}})$ for some $x_{\textit{min}}\in K$, so I cannot (at least directly) say that $(y_j)$ is the sequence of minimizers of the $(f_j)$
$\inf_{x \in K} f_j(x)+\frac 1j \geq f_j(y_j)$ for some $y_j \in K$. Hence, $f_j(y_j) +\frac 1 j\geq \inf_{x \in K} f_j(x)+\frac 1j \geq f_j(y_j)$ Take $\lim \inf$.