What’s fishy in the proof of the theorem “Every closed subset of a compact set is compact.”

Question

Let $X$ be a metric space. Let $E\subset X$ be a compact subset of $X$ and let $F\subset E$ be a closed subset of $E$.

This is how I tried to prove the theorem:

Suppose, on the contrary that $F$ is not compact. This implies that no open cover of $F$ contains a finite subcover.

Since, $E$ is compact relative to $X$, there exist finite no. of open sets (in $X$) $V_1,V_2,…V_n$ such that
$E\subset V_1\cup V_2\cup \cdots\cup V_n$
Since, $F\subset E$, we have $F\subset V_1\cup V_2\cup \cdots\cup V_n\implies F$ is compact relative to $X $ $\tag{1}$.
Now, it is known that if $F\subset E\subset X$, then $F$ is compact relative to $X$ if and only if $F$ is compact relative to $E$.
Hence by $(1)$: $ F$ is compact relative to $E$, which is a contradiction.

Has something gone wrong in this proof? I think so because no where have I used the fact that "$F$ is a closed subset." Please help. Thanks.

Answer 1

Where your proof went wrong is violating the definition of compactness of $E$ relative to $X$. This does NOT mean

... there exist finite no. of open sets (in $X$) $V_1,V_2,...V_n$ such that $$E\subset V_1\cup V_2\cup \cdots\cup V_n$$

Instead, what compactness of $E$ relative to $X$ means is

... for every collection of open sets $\{V_i\}_{i \in I}$ (in $X$) such that $$E \subset \bigcup_{i \in I} V_i $$ there exists a finite sub collection $V_{i_1},V_{i_2},...,V_{i_n}$ such that $$E \subset V_{i_1} \cup V_{i_2} \cup \cdots \cup V_{i_N} $$