Why are subsets of compact sets not compact?
There's a nice symmetry/duality here stemming from the fact that "finite" has two generalizations in the topological category.
Images and subsets of finite sets are finite.
Discrete topological spaces generalize finite sets in that subspaces inherit the property.
Compact topological spaces generalize finite sets in that images inherit the property.
A topological space that is discrete and compact is finite.
One way to think of it is that compactness means that sequences (more generally, nets) cannot "run away". There are two types of "running away":
- Running away to infinity (failure of the set to be bounded, or some analog*. Failure of the sequence/net to have any limit point at all).
- Running away to a limit point which isn't in the set (failure of a set to be closed. The sequence/net has a limit point(s) in an ambient space, but that point is missing from the set in question).
If I have a compact space $X$, and I remove a point, $X-\{p\}$ may suddenly permit the second type of "running away".
At first it seemed awkward to me to phrase these things in terms of sequences and nets, because sequences and nets are "discrete" objects describing a continuous thing. But one can always phrase everything in terms of nets/filters of open sets, not of points. That can make it seem a bit more natural.
In any case, the basic point of a compact set is that it does not allow you to play a certain kind of game with infinity.
*For example, a uniform space is compact iff totally bounded and Cauchy complete, which are exactly analogous to conditions 1 and 2 above.
What you may be after is the idea of a "pre-compact" subset, that is, a subset whose closure is compact. Then every subset of a pre-compact set is pre-compact; as are finite unions of them. A compact set is a closed pre-compact set.
For metric spaces, there is a closely connected concept of "totally bounded" subsets; a complete totally bounded set is compact.
Because a subset of a compact set is smaller than the compact set, the subset might have a different open cover that does not cover the compact set.
This open cover may not have a finite subcover.
Example: Let $A = [0,1]$ and $B = (0, 1] \subset A$.
Now $U = \{U_i| U_i= (\frac 1i, 1.1)\}$ is an open cover of $B$ but it is not an open cover of $A$. (And $U$ does not have a finite subcovering of $A$.)
To extend $U$ so that that it will cover $A$ we must add an open set that contains $0$. Call that $U_{\alpha}$ and $0 \in U_{\alpha}$ and now $U \cup \{U_{\alpha}\}$ is an open cover of $A$.
But $U_{\alpha}$ is open so there is an $r > 0$ so that $N_r(0) = (-r, r) \subset U_\alpha$. But we can find an $n > \frac 1r$ or in other words $0 < \frac 1n < r$.
So $(0, \frac 1n] \subset U_{\alpha}$ so $(0, \frac 1n]$ is covered but the single open set $U_{\alpha}$. Without $U_{\alpha}$ and with only $U = \{U_i = (\frac 1i, 1.1)\}$ we would have needed an infinite number of $U_i| i > n$ to cover $(0, \frac 1n]$. But with $U_{\alpha}$ we don't need ANY of them anymore.
So ... throw them away! We are left with $\{U_\alpha\} \cup \{U_i|i \le n; n > \frac 1r\}$ and that is a finite subcover of $A$. And of $B$.
But the point is. Without the requirement that there is an open set containing $0$ we wouldn't have a situation where a single open set must "do the work" of an infinite number of open sets which a non-compact set without the point $0$ could require.
... more explicitly with maybe too much detail...
So $A \setminus U_\alpha \subset (\frac 1n, 1]\subset B$. And $(\frac 1n, 1]$ is covered by the finite subclass $\{U_i| i \le n\}$ and we don't need $\{U_i| i > n\}$ any more because $U_\alpha$ covers everything in $A$ that was not covered in $\{U_i|i > n\}$.
(Namely $U_{\alpha}$ covers $\{0\} \cup (0, \frac 1n]$ whereas without $U_\alpha$ we needed ALL of $\{U_i| i > n\}$ to cover $(0, \frac 1n]$)
So $\{U_i|i \le n\} \cup \{U_\alpha\} \subset U \cup \{U_\alpha\}$ is a finite subcover of $A$. (even that $U$ had no finite subcover of $B$.