What's fishy in my proof of the theorem "Every closed subset of a compact set is compact."? 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.
 A: 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
$$E \subset V_{i_1} \cup V_{i_2} \cup \cdots \cup V_{i_N}
$$

A: You write

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

This is wrong. If $F$ is not compact, then there exists at least one open cover of $F$ that doesn't contain a finite subcover.
Just think to $X$, which provides an open cover of $F$ and is finite.
You go on with another incorrect statement. If $E$ is compact, not only it is the covered by a finite number of open sets (this is trivial, just take $X$ itself), but every open cover of $E$ contains a finite subcover.
This is the crux of compactness: every open cover contains a finite subcover, the negation of which is there is an open cover that contains no finite subcover.

The proof is actually quite simple and doesn't require contradiction. Take an open cover $\mathcal{V}$ of $F$. Since $F$ is closed in $E$, there exists a closed set $C$ in $X$ such that $F=E\cap C$. Now add $V=X\setminus C$, which is an open set in $X$. Then $\mathcal{V}\cup\{V\}$ is an open cover of $E$, so you can find $V_1,V_2,\dots,V_n\in\mathcal{V}$ such that
$$
E\subseteq V\cup V_1\cup V_2\cup\dots\cup V_n
$$
Therefore $F\subseteq V_1\cup V_2\cup\dots\cup V_n$.
Remark 1. Maybe it's not required to use $V$ in the finite subcover, but adding it does not change finiteness and allows to finish the proof.
Remark 2. It's actually not restrictive to assume $E=X$, because compactness is an “absolute” property: it doesn't depend on what space $E$ is embedded in. Proving this useful property is an interesting exercise.
Remark 3. You could go by contradiction. Suppose there is an open cover $\mathcal{V}$ of $F$ that contains no finite subcover. Then, using the same symbols as before, $\mathcal{V}\cup\{V\}$ is an open cover of $E$ that cannot contain a finite subcover, because this would also be a finite subcover of $F$.
