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I have encountered a characterization of compact space, but I do not know how to prove it.

$V$ is a topological space which satisfies that for any topological space $W$, the projection $V\times W\rightarrow W$ is a closed map, then $V$ is compact.

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Your question is equivalent to the statement that the function from $V$ to a one point space is a proper map. Read more about proper maps in Bourbaki's general topology, for example. – commenter Nov 1 '12 at 14:31

Let $\phi$ be a filter on $V$. Let $W = V \cup \{\phi\}$, where $V$ carries its topology and $F \cup \{\phi\}$, $F \in \phi$ are the neighbourhoods of $\phi$.

Now let $D = \{(x,x)\mid x \in V\}$ and $A = \operatorname{cl}D$. Now $\operatorname{pr}_W(A)$ is a closed subset of $W$ containing $\operatorname{pr}_W(D) = V$. Hence $\operatorname{pr}_W(A) = W$, so there is a $x \in V$ such that $(x, \phi) \in \operatorname{cl}D$, that is each neighbourhood $U \times (\{\phi\} \cup F)$ of $(x, \phi)$ intersects $D$, that is $U \cap (\{\phi\} \cup F) = U \cap F\ne\emptyset$ for each $F \in \phi$, $U$ neighbourhood of $x$. So $x$ is an adherence point of $F$ and $V$ is compact.

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I have a question: How do we know that the neighborhood base $\{F\cup\{\phi\}\mid F\in\phi\}$ of $\phi$ is compatible with the topology on $V$ ? Would we have to show that each $F$ contains a $G\in\phi$ such that each $x\in G$ is interior point of $F$? – Stefan Hamcke Jul 6 '15 at 16:00

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