Isn't every subset of a compact space compact? Let $\mathfrak{C}$ be an open cover of a topological compact space $X$, and let $\mathfrak{B}\subseteq \mathfrak{C}$ be its finite subcover. Then every subset of $X$ also has the same cover and subcover! Shouldn't every such subset also be compact then?
Motivation: I read somewhere that every closed subset of a compact space is compact. I wonder why it shouldn't be the case for every subset of $X$, rather than just the closed ones. 
 A: Not every open cover of subset of $X$ is cover of $X$. 
For example, let $X=[0, 1]$. Then $(0, 1)$ is a subset of $X$.
If we take $A_n =(1/n, 1),\, n=2, 3, \ldots$, then $A_n,\, n=2, 3, 4, \ldots$
is a cover of $(0, 1)$ but not of $X$. 
A: The big issues here are


*

*That open sets of a subspace are not always open in the larger space.

*That the open sets you add to get a cover of the whole space may have to be very large and contain infinitely many of the subspaces open sets. These large open sets are often preferentially chosen in the finite subcover, like in the one point compactification of a discrete space.

A: 
Then every subset of $X$ also has the same cover and subcover!

Every subset of $X$ has the same cover and subcover. But there are covers of the subset that would not cover $X$. This cover might not have a subcover. The examples are stated above.
A: The point is that EVERY cover must admit a finite subcover, and only because this is true for the ambient space, does not mean it is true for every subset.
So given a compact space $X$ and $U \subset X$ arbitrary, you start with an open cover of $U$, and try to find a finite subcover of it.
But not every cover of $U$ comes from a cover of $X$, not even if $U \subset X$ is open.  Consider $X=[0,1] \subset \Bbb{R}$, and $U=(0,1) \subset [0,1]$ and the cover 
$$
U=\bigcup_{i=1}^{\infty}\left(\frac{1}{n},1-\frac{1}{n} \right)
$$
There is no open cover of $[0,1]$ which restricts to the above cover on $(0,1)$.
