Let $V$ be the set of truth valuations in propositional logic, and let $\Delta \subseteq PROP$. We define $$[\Delta] = \{v \in V : v(\Delta)=1\}$$. (If $\Delta =\varphi$, then we can write $[\Delta] =[\varphi]$).
I already know that the set $$B =\{ [\varphi] : \varphi \in PROP \} $$ form a basis for a topology $\tau$ on $V$, and that the sets of the form $[\varphi]$ are clopen in $\tau$.
The problem is: prove that the Compactness theorem for propositional logic is equivalent to the topological compactness of $(V,\tau)$.
There're also two hints: (1) prove that the closed sets are exactly those of the form $[\Delta]$; (2) use the characterisation of topological compactness given in terms of closed sets.
Let $\Delta$ be a set of propositional formulae. We show that $V \setminus [\Delta]$ is open. In fact, $V \setminus [\Delta]$ may be expressed as the union of all $[\neg \phi]$ where $\phi$ belongs to $\Delta$: if $v$ is a valuation belonging to $[\neg \phi]$, then $v(\phi) = 0$ and so $v \notin [\Delta]$. Also, if $v \in V \setminus [\Delta]$, $v$ evaluates at least one formula $\phi \in \Delta$ as 0, and so $v$ belongs to $[\neg \phi]$. The sets $[\neg \phi]$, where $\phi \in \Delta$, are all open, so their union is open.
In the other direction, we have to show that if $X$ is a closed set of $V$, $X$ is of the form $[\Delta]$. $V \setminus X$ is open and is thus the union of a family of sets, each of which is $[\phi]$ for some formula $\phi$. (Here we have used the given base for $V$.) In other words, there is a set $\Gamma$ of formulae such that $$V \setminus X = \bigcup_{\phi \in \Gamma} [\phi]$$ Define $$\Delta = \{\neg \psi : \psi \in \Gamma\}$$ We claim that $[\Delta] = X$: if $v \in [\Delta]$, $v$ evaluates all formulae $\phi$ of $\Gamma$ as 0. Therefore $v \notin V \setminus X$. If $v \in X$, $v$ does not belong to any set $[\phi]$ for a formula of $\Gamma$. Therefore $v$ evaluates all formulae of $\Gamma$ as 0, so $v \in [\Delta]$.
This characterisation is: every family of closed subsets of $V$ which satisfies the finite intersection property has a non-empty intersection. If $\Gamma$ is a set of formulae such that every finite subset of the family has a valuation which makes them all true, then $$A = \{[\phi]: \phi \in \Gamma\}$$ is a family of closed subsets with the finite intersection property. The intersection of $A$, which, we know by compactness, is not empty, must therefore contain a valuation which satisfies all of $\Gamma$. Now that you have the idea, it should be straightforward to prove the other direction, namely that the compactness of propositional logic implies that $V$ is a compact space.