universal nets in compact spaces are convergent

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

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

$${\forall }_{\alpha \in 𝒜}{\exists }_{\alpha \le {\alpha }_0}{x}_{{\alpha }_0}\notin U_x$$

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

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

$${\forall }_{{\alpha }_k\le \alpha }{x}_{\alpha }\in X-U_{x_k}$$

Because we have a finite number ${\alpha }_1,{\alpha }_2\mathrm{\dots },{\alpha }_n\in 𝒜$ we can find $\gamma \in 𝒜$ such that ${\alpha }_k\le \gamma$ for each $1\le k\le n$.

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