Some elementary results on compact spaces, a couple of simple proofs, and one deeper result.
Lemma Let X be a compact topological space and let C be a closed subset of X. Then C is compact.
Proof Let A be an open cover of C.
Since C is closed in X, the set X\C is open in C, and so A union {X\C} is an open cover of X. Since X is compact, this has a finite subcover B of X, and then B\{X\C} is a finite open cover of C which is a subcover of A. So C is compact.
Lemma A compact subspace of a Hausdorff topological space X is closed in X.
Proof Let x be in X\C. X is Hausdorff, so for each c in C there are disjoint open sets Uc and Vc such that Uc contains c and Vc contains c.
Then the collection {Uc : c in C} is an open cover of C. Since C is compact, this has a finite subcover {Uc1, Uc2, ..., Uck}.
Now let Vx = Intersect{Vc1, Vc2, ..., Vck}. Vx is the intersection of finitely many open sets containing x, and so is an open set containing x. Moreover, Vx does not intersect C; for if y is in C then y is in Ucr for some r, and so y is not a member of Vcr, and hence is not in Vx.
What we have shown so far is that whenever x is in X\C then there is an open set Vx which contains x and does not meet C. Then Union{Vx : x in X\C} is open, contains X\C, and does not meet C, so in fact it must equal X\C.
Then X\C is open, so C is closed.
Lemma Let X,Y be topological spaces with X compact and let f:X->Y be continuous and onto. Then Y is compact.
Lemma A topological space X is compact if and only if the following property holds: whenever Z is a collection of closed sets in X such that the intersection of finitely many elements of Z is nonempty, then Intersection(Z) is nonempty.
Proof Repeated application of De Morgan's laws for sets.
Lemma A compact subset of a metric space is bounded.
Proof Let (X,d) be a metric space, C a compact subset.
If Y is empty, then Y is vacuously bounded. Otherwise let y be in Y, and let Un = {y' in Y : d(y,y')<n}. Then {Un : n in N} is an open cover of Y, and so it has a finite subcover {Un1, Un2, ..., Unr}. Then d(y,y')<r whenever y' is in Y, so Y is bounded.
Lemma A finite product of compact spaces is compact.
While the last result can be proven using elementary methods, the generalisation to infinite products is much harder to prove, and is in fact equivalent to the axiom of choice.
Theorem (Tychonoff) An arbitrary product of compact spaces is compact in the product topology.