${\limsup\limits_{n\to \infty}}(\inf_\limits{x \in X}{f_{n}(x)}) \leqslant \inf_\limits{x \in X} ({\limsup\limits_{n\to \infty}} f_{n}(x))$

$X \subset \mathbb{R}$ and a function is defined as such $f_{n} : X \to \mathbb{R}, n \in \mathbb{N}$ so that a set $\left\{ f_{n}(x): n \in \mathbb{N}, x \in X \right\}$ is unbounded.

Prove that

${\limsup\limits_{n\to \infty}}\left(\inf_\limits{x \in X}{f_{n}(x)}\right) \leqslant \inf_\limits{x \in X} ({\limsup\limits_{n\to \infty}} f_{n}(x))$

I was thinking something like this

$f_{n}(x)≤\sup_\limits{x\in X}f_{n}(x)$

$⇒ \liminf_\limits{n→\infty} f_{n}(x) ≤ \liminf_\limits{n→∞} (\sup_\limits{x∈X}f_{n}(x))$

$⇒ \sup_\limits{x∈X} (\liminf_\limits{n→∞}f_{n}(x)) ≤ \liminf_\limits{n→∞}( \sup_\limits{x∈X} f_{n}(x))$

But the end result isn't quite what I needed.

For any $x\in X$, $$\inf_{x'\in X} f_n\left(x'\right)\leqslant f_n(x),$$ hence $$\limsup_{n\to +\infty} \inf_{x'\in X} f_n\left(x'\right)\leqslant \limsup_{n\to +\infty}f_n(x).$$ Since $x$ is arbitrary, it follows that $$\limsup_{n\to +\infty} \inf_{x'\in X} f_n\left(x'\right)\leqslant \inf_{x\in X} \limsup_{n\to +\infty}f_n(x).$$