Every subspace of a compact space is compact? It seems as if every subspace of a compact topological space (equipped with its relative topology) had ought to be compact as well. Is this true in general?
And in particular, I want to use the fact that every quasi-projective variety is compact with the Zariski topology - I remember having proven it for quasi-affine varieties, but the same proof does not work for quasi-projective varieties. Is every quasi-projective variety compact?
 A: A topological space $X$ is called quasi-compact if every open covering of $X$ has a finite subcovering.
(This is the terminology adopted by algebraic geometers to emphasize that Hausdorffness is not required. Compact spaces are then Hausdorff quasi-compact spaces).     
A topological space $X$ is said to be noetherian if every non-empty family of closed subspaces has a minimal element.
For example  $\mathbb A^n_k$ and $\mathbb P^n_k$     are noetherian: this follows from Hilbert's theorem that the polynomial ring $k[T_i,...,T_n]$ is noetherian .
The following are then equivalent for a topological space $X$:
$\bullet $ $X$ is noetherian.
$\bullet $  Every subset of $X$ is quasi-compact.     
Since  projective space $\mathbb P^n_k$ over a field  is noetherian ,  any quasi-projective variety is quasi-compact because by definition it is  the intersection of an open and a closed subset of some  $\mathbb P^n_k$, and any subset of   $\mathbb P^n_k$ is quasi-compact by noetherianity of  $\mathbb P^n_k$.  
To sum-up:
Every quasi-projective variety is quasi-compact.     
Edit: as emphasized by Pete, even an arbitrary   subset of a quasi-projective variety is quasi-compact, since it too is a subset of  $\mathbb P^n_k$.
A: A closed subspace $A$ of a compact space $X$ is compact.
Proof (Pardon my terribly sloppy indexing): Take an open cover $\{U_\alpha \cap A\}_{\alpha}$ of $A$ where each $U_\alpha$ is open in $X$. Then $\{U_\alpha\}_{\alpha} \cup \{X \setminus A\}$ is an open cover of $X$, which has a finite subcover. The $U_\alpha$ in this finite subcover necessarily cover $A$. So finitely many $\beta$ can be chosen such that the collection of these $U_\beta \cap A$ cover $A$.
A: A space such that all of its subspaces are compact is hereditarily compact. This property turns out to be implied by being Noetherian, which holds in particular for the projective spaces $\mathbb{P}^n$. 
A: Every closed subspace of a compact Hausdorff space is compact.  I'm not actually sure whether that applies to non-Hausdorff spaces; I probably used to know.
But $[0,1]$ is compact, and $(0,1)$ is a non-compact subspace.
If what you propose were true, then one would not speak of "compactifications".  A compactification of a topological space $X$ is a space $Y$ such that $X\subseteq Y$ and the closure of $X$ in $Y$ is precisely $Y$.  For example, if one adjoins a single point $\infty$ to the real line $\mathbb{R}$ in such a way that an open neighborhood of $\infty$ is any set of the form
$$\{\infty\}\cup \{x\in\mathbb{R} : x > a\}\cup \{x\in\mathbb{R} : x < b\},$$
then that is the "one-point compactification" of $\mathbb{R}$.  The line also has a two-point compacification.  One can show that it has no three-point compactification, but it has some much larger compactifications with infinitely many points.
A: You can surely answer this yourself! It is a matter of considering examples.
Let us say you start with the very first topological space you were introduced to: the space of real numbers. It is not compact, but it has lots of compact subspaces, and in fact we can describe them all. Pick one and look at its subspaces now...
