Theorem - A universal net $ (x_{\alpha})_{\alpha \in \mathcal{A}}$ in a compact space $ X$ is convergent.

Proof : Suppose by contradiction that $(x_{\alpha})_{\alpha \in \mathcal{A}}$ was not convergent. Then for every $x \in X$ we would find neighborhoods $U_x$ such that $$ \forall_{\alpha \in \mathcal{A}}\;\;\; \exists_{\alpha \leq \alpha_0} \;\;\; x_{\alpha_0} \notin U_x $$

The collection of all this neighborhoods cover $X$ , and as $X$ is compact, a finite number $U_{x_1}, U_{x_2}, \dots, U_{x_n}$ also cover $X$ .

The net $(x_{\alpha})_{\alpha \in \mathcal{A}}$ is not eventually in $U_{x_k}$ so it must be eventually in $X-U_{x_k}$ (because it is a universal net). Therefore we can find $\alpha_k \in \mathcal{A}$ such that $$ \forall_{\alpha_k \leq \alpha} \;\;\; x_{\alpha} \in X-U_{x_k} $$

Because we have a finite number $\alpha_1, \alpha_2 \dots, \alpha_n \in \mathcal{A}$ we can find $\gamma \in \mathcal{A}$ such that $\alpha_k \leq \gamma$ for each $1 \leq k \leq n$ .

Then $x_{\gamma} \in X-U_{x_k}$ for all $k$ , i.e. $x_{\gamma} \notin U_{x_k}$ for all $k$ . But $U_{x_1}, U_{x_2}, \dots, U_{x_n}$ cover $X$ and thus we have a contradiction. $\square$