Question about compactness and regular spaces I was reading part of an article of S.H. Hechler, "On some weakly compact spaces". In the proof of theorem 3.1, he states the following:
"We have, therefore, shown that every open cover of [the space] $T$ admits a finite subcover of a [countable] dense subset $D$, and it is well known that in regular spaces this implies compactness."
That is not clear for me. Can someone explain me why or give me a reference?
Question: I must show that if every open cover of $T$ admits a finite subcover of a dense subset $D$, then, supposing $T$ is regular, $T$ is compact.

Let $(O_i)_{i\in I}$ be any open cover of $T$. Since $T$ is regular, one can find an open cover $(V_j)_{j\in J}$ such that, for each $j\in J$, the closure $\overline V_j$ of $V_j$ is contained in $O_{i(j)}$, for some $i(j)\in I$. Next, by assumption on $T$, one can find $j_1, \dots ,j_N\in J$ such that $V_{j_1}\cup\dots \cup V_{j_N}$ is dense in $T$. Then $T=\overline{V_{j_1}\cup\dots \cup V_{j_N}}=\overline{V_{j_1}}\cup\dots \cup\overline{V_{j_N}}\subset O_{i(j_1)}\cup\dots\cup O_{i(j_N)}$, and hence $(O_{i(j_1)},\dots ,O_{i(j_N)})$ is a finite subcover of the original cover$(O_i)_{i\in I}$.
