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To see that $S$ is countably compact, suppose that we have a countable open cover $\mathcal{U}$. Notice that $S$ is the union of the subspaces $$(\omega_1+1)\times (\alpha+1)\cup (\alpha+1)\times(\omega_1+1)$$ for $\alpha<\omega_1$. These are the L-shaped parts of the plane you have defined, proceeding from the origin at lower left. Being the union of two products of compact spaces, each of these spaces is compact in the subspace topology. So for each $\alpha$, there is some finite subcover of $\mathcal{U}$ covering the $\alpha$th subspace. And since the subspaces are accumulating, this cover also covers all the earlier subspaces. But there are only countably many finite subsets of $\mathcal{U}$, so some particular finite subfamily must have been used for unboundedly many $\alpha$, and so this particular subcover covers all the subspaces, and hence covers $S$. So $S$ is countably compact.