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Let $X$ be the set of all real numbers with the co-countable topology. It is easy to show that a compact subset of $X$ is finite.

Is it true that every compact subset of $X\times X$ also finite? I guess yes. If $K$ is a compact subset of $X\times X$ then there is compact $A \subseteq X$ such that $K \subseteq A\times A$. Since $A$ is finite, $K$ is also finite.

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    interesting fact: a space where the only compact subsets are the finite ones (which regardless of the space are *always* compact), is called "anti-compact"; there is a whole litle theory on anti-properties (from Ireland mostly). – 2017-02-04
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    So the argument shows that the product of finitely many anti-compact spaces is anti-compact. – 2017-02-04

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You're quite right. If $K \subset X \times X$ is compact, then $\pi_1[K]$ is compact in $X$ so finite. Also, $\pi_2[K]$ is compact in $X$, so finite, and $K \subseteq \pi_1[K] \times \pi_2[K]$ and the right hand set is finite too. Note that "the continuous image of a compact subspace is compact" needs no separation axioms on the spaces involved.