Lemma 5.12.3. A closed subset of a quasi-compact topological space is quasi-compact.

Proof. Let $E \subset X$ be a closed subset of the quasi-compact space $X$. Let $E = \bigcup V_ j$ be an open covering. Choose $U_ j \subset X$ open such that $V_ j = E \cap U_ j$. Then $X = (X \setminus E) \cup \bigcup U_ j$ is an open covering of $X$. Hence $X = (X \setminus E) \cup U_{j_1} \cup \ldots \cup U_{j_ n}$ for some $n$ and indices $j_ i$. Thus $E = V_{j_1} \cup \ldots \cup V_{j_ n}$ as desired. $\square$

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