Let $X$ be a closed subset of a compact metric space $M$. Then, $X$ is compact. Theorem : Let $X$ be a closed subset of a compact metric space $M$. Then, $X$ is compact.
Query : Since, $M$ is compact, then there is a finite collection $F$ of open sets which covers $M$. Hence, $F$ should cover every subset of $M$, whether be it open or closed. 
Then, why does the problem define this property only for closed subsets of the compact metric?
Thank you for your help.
 A: You misunderstand the definition of compactness of $X$. This says that for every open cover of $X$ we can find a finite subcover that also covers $X$. This reduction has to be possible for every open cover; every space can be covered by just one open set, namely itself, so having a finite cover means nothing at all. 
So to your problem: You know $M$ is compact, and $X \subset M$ is closed. To show compactness, start by taking any open cover of $X$. To use compactness of $M$, can you augment this cover of $X$ to one of $M$ so that a finite cover of the larger cover also gives you a finite cover for the original? One that uses the closedness?
A: General version: a closed subspace $X$ of a compact topological space $M$ is compact.
Proof: Suppose we have a covering $$X = \bigcup_i U_i$$ with $U_i$ open in $X$. Then there exist $A_i $ open in $M$ such that $U_i = X \cap A_i \ \forall \ i $ . 
Moreover we have $$ M = (M \setminus X) \bigcup ( \bigcup_i A_i ) $$ M is compact so $$M = (M \setminus X) \bigcup ( \bigcup_{j = 1}^{n} A_{i_j} )  $$ for some $i_j$. This implies $$X = \bigcup_{j = 1}^{n} U_{i_j} $$ and so $X$ is compact. 
A: Since $M$ is compact, for any sequence ${x_n} \ \ s.t. x_n\in X$, there exists a convergent subsequence ${x_n}_k$ of ${x_n} \ \ s.t. \ \ {x_n}_k \to x$, for some $x$ in $M$. Clearly, $x$ is a limit point of $X$, and thus by closeness of $X$, $x \in X$.
Since any sequence in $X$ has a convergent subsequence in $X$, $X$ is compact as a result. 
A: Because it's false for open subsets. For instance, $(0,1)$ is not compact.
The proof:
If $F\subset X$ is a closed subspace of $X$, given $A_1,\ldots,A_i, \ldots$ open sets of $F$ such that $F = \cup A_i$, then $X = \cup A_i \cup (X\setminus F) $, but each $A_i = F \cap O_i$, where $O_i$ is a open subset of $X$. Then $X = \cup O_i \cup (X\setminus F)$, hence there is a finite sub-colection $\mathcal{F}$ of $\{X\setminus F, O_1, \ldots, O_n, \ldots\}$ such that $X = \bigcup \mathcal{F}$
Define $\mathcal{G} = \{ F \cap Y : Y \in \mathcal{F}\}$, then it's clear that $F = \bigcup \mathcal{G}$ and $\mathcal{G}$ is a finite sub-colection of $\{A_1,\ldots,A_i, \ldots\}$.
QED
Can you see what fails if $F$ is open? $X\setminus O$ won't be open and we can't use the compacity o $X$ in our demonstration.
A: I think I see the confusion here: Let $\bigcup O$ be the union of any collection of open sets such that $M \subseteq \bigcup O$. Then there is a finite subcover of $\bigcup O$ that covers $M$. The key word here is any.
Then as you noted, let $K$ be any closed set. Since $K \subseteq M$, $K$ must also be covered by the finite subcover.
Now, if $H$ is an open subset of $M$, it is still a subset of the finite open cover of $M$. However, by the definition of a compact set, there is always some new collection $O'$ which, if we define it cleverly enough, $H$ does not have a finite subcover.
So, even if an open set is a subset of some finite subcover, there is always another cover from which we cannot obtain a finite subcover.

Example: Consider $(0,1)$. Let $O_x = (x/2,1)$. Then $\bigcup_{x\in (0,1)}O_x$ is a cover, but as an exercise you can prove that there is no finite open subcover. 
