Weak compact implies weak* compact? Let $S$ be a set which is compact in the weak topology. Is it true that $S$ is also compact in the weak* topology?
I mean, the weak* topology is coarser than the weak topology, but the existence of the first finite subcover does really imply another finite subcover? Can we take, for each open set $U$ another open set $U'$ containing $U$?
More generally, if $S$ is compact in a topology $\mathcal{T}$, then is it also compact in the topology $\mathcal{T}' \subseteq \mathcal{T}$?
 A: Yes. If $\mathscr{U}\subseteq\mathcal{T}'$ such that $\mathscr{U}$ covers $S$, then as $\mathscr{U}$ is also a $\mathcal{T}$-open cover of $S$ and $S$ is $\mathcal{T}$-compact, we obtain $\mathscr{U}_0\subseteq \mathscr{U}$ finite such that $\mathscr{U}_0$ covers $S$. Thus we have found a finite subcover.
Interestingly enough, we get a dual analogue for Hausdorff - and the two come together nicely.

Let $\tau_C$, $\tau_H$, and $\tau$ be topologies on a set $X$ such that $(X,\tau_C)$ is compact and $(X,\tau_H)$ is Hausdorff.
(i) If $\tau\subseteq\tau_C$, then $(X,\tau)$ is compact.
(ii) If $\tau\supseteq\tau_H$, then $(X,\tau)$ is Hausdorff.
(iii) If $\tau_H \subseteq \tau_C$, then $\tau_C=\tau_H$.

The proof of (i) is above and the proof of (ii) is quite simple. For (iii), suppose $f:(X,\tau_C)\to (X,\tau_H)$ is the identity function on $X$. From $\tau_H\subseteq\tau_C$, we deduce that $f$ is continuous. Then $f$ is a continuous bijection from a compact space into a Hausdorff space, which means that $f^{-1}$ is continuous. Therefore $\tau_C\subseteq\tau_H$.
A: A set is compact iff every open cover has a finite subcover. So if we throw away open sets (make it coarser) there are only fewer open covers of $X$ to "test" compactness with. Any open cover in the coarser weak$^\ast$ topology is in particular an open cover in the weak topology. So we can find a finite subcover by virtue of its compactness in the (finer) weak topology. (A subcover means that the sets we use are still the same ones, but just fewer of them; it could be different with other properties of the topology, like connectedness or normality). 
So if you can do the "compactness test" for all open covers from a topology, then you can certainly do it for the subfamily of covers from the weaker topology. The fewer open sets there are, the easier it is to be compact:
Extreme example: in the indiscrete topology $\{\emptyset, X\}$ on $X$ all subsets of $X$ are compact, in the dicrete topology, only the finite ones.
