Why isn't every subset of a compact set also compact? I've seen on MathSE that a subset of a compact set need not be compact. However, I am not fully understanding why. 
If $X$ is compact then there exists a finite subcover, i.e., $X \subseteq \bigcup_{i=1}^n U_i$. So any subset of $X$, say $K$, will be a subset of the finite union of $U_i$.
So why is not the case that $K$ is automatically compact?
 A: An open cover of $K \subseteq X$ may not be an open cover of $X$. Indeed, for $K$ to be compact, you need to start with an open cover of $K$, not of $X$, and find a finite subcover.
A: Not every open cover of $K$ is an open cover of $X$. For instance, if $X=[0,1]$ and $K=(0,1)$ then $U_n=(1/n,1-1/n)$ is an open cover of $K$ but not $X$.
A: What you've shown is that any open cover $\mathcal{A}$ of $K$ has a finite subcover . . . if $\mathcal{A}$ is also a cover of $X$! But that need not happen. For a concrete example, consider $K=(0, 1)$ as a subset of $X=[0, 1]$. Then $\mathcal{A}=\{({1\over n}, 1-{1\over n}): n\in\mathbb{N}\}$ is a cover of $K$ with no finite subcover. Note that $\mathcal{A}$ fails to cover $X$, but there was no reason for it to cover $X$ in the first place.
A: "Compactness" is a finiteness condition for topological spaces which is infinitely more subtle than we are able to imagine. Note that the open unit disk $D\subset{\mathbb R}^2$, a standard metric space, would be compact according to your definition, since it is a subset of the compact square $Q:=[{-1},1]^2$, but you'll have to admit that $D$, considered as a subspace of ${\mathbb R}^2$ is not compact: The sequence ${\bf z}_n:=\bigl(1-{1\over n},0\bigr)$ $(n\geq1)$ in $D$ has no accumulation point in $D$.
