$S$ is compact if every net in $S$ has a convergent subnet

Question

I am going through a proof in Reed & Simon's book on functional analysis that states a space $S$ is compact if every net in $S$ has a convergent subnet. Their proof is:

Suppose that every net has a convergent subnet and let $\mathcal{U}$ be an open cover. Let us suppose that $\mathcal{U}$ has no finite subcover and derive a contradiction. Order the finite subfamilies $\mathcal{G}$ of $\mathcal{U}$ by inclusion; $\mathcal{G}$ is thereby a directed set. For each $\mathcal{F} = \{F_1, \ldots, F_m\} \in \mathcal{G}$, pick $x_\mathcal{F} \not\in \bigcup_{i=1}^m F_i$. By assumption, the net $x_\mathcal{F}$ has a cluster point $x$. Since $\mathcal{U}$ is a cover, we can find $U \in \mathcal{U}$ with $x \in U$. Since $x_{\mathcal{F}}$ is frequently in $U$ we can find a finite subfamily $\mathscr{G} \in \mathcal{G}$ so that $\{U\}\prec \mathscr{G}$ and $x_\mathscr{G} \in U$. Since $\{U\} \prec \mathscr{G}$, $U \subset \bigcup_{G \in \mathscr{G}} G$, and so $x_{\mathscr{G}} \in \bigcup_{G \in \mathscr{G}} G$, which is a contradiction.

I have a few questions about this proof which has been causing me some trouble.

1. How does the assumption that every net has a convergent subnet imply that the net $\{x_\mathcal{F}\}$ has a cluster point $x$? Or does this follow from some other assumption?
2. I am confused on what is going on in the step where they say:

Since $x_{\mathcal{F}}$ is frequently in $U$ we can find a finite subfamily $\mathscr{G} \in \mathcal{G}$ so that $\{U\}\prec \mathscr{G}$ and $x_\mathscr{G} \in U$.

Are they finding a subnet of $\{x_\mathcal{F}\}$, denoted $\{x_{\mathscr{G}}\}$, whose limit point we know exists by assumption and which we call $x_\mathscr{G}$?

3. Where is the contradiction?

Answer 1

1. The net $(x_\mathcal{F})_{\mathcal F\in\mathcal G}$ has a subnet converging to some $x,$ which is then a cluster point of the net.
2. Since $x$ is a cluster point of $(x_\mathcal{F})_{\mathcal F\in\mathcal G}$ and $U$ is a neighborhood of $x,$ for every $\mathcal F_0\in\mathcal G,$ there exists an $\mathcal F\in\mathcal G$ such that $\mathcal F_0\subset\mathcal F$ and $x_{\mathcal F}\in U.$ Apply this to $\mathcal F_0=\{U\}.$
3. $x_{\mathscr{G}} \in \bigcup_{G \in \mathscr{G}} G$ contradicts the definition of the net $(x_\mathcal{F})_{\mathcal F\in\mathcal G},$ where each $x_{\mathcal F}$ was "picked" outside $\bigcup_{F\in \mathcal F}F.$