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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 nested chain of 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.

Terminology. Let me also mention that in my experience what is ordinarily called the Tychonoff plank is a bit different from your space. For example, on the Wikipedia entry Tychonoff plank, the Tychonoff plank is defined to be the space $$(\omega_1+1)\times(\omega+1)$$ and the deleted Tychonoff plank is $$(\omega_1+1)\times(\omega+1)\setminus\{(\omega_1,\omega)\}.$$ These differ from yours in that the vertical dimension reaches only up to $\omega$ rather than $\omega_1$, and so it is a "plank" in being much longer than it is wide. Your space is more like a quarter plane than a plank.

But let me point out that the deleted Tychonoff plank as defined on Wikipedia is not countably compact, since we can take the open set $\omega_1\times(\omega+1)$ together with the horizontal lines $(\omega_1+1)\times\{n\}$. This is a countable cover of the deleted plank, but there is no finite subcover.

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 nested chain of 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.

Terminology. Let me also mention that in my experience what is ordinarily called the Tychonoff plank is a bit different from your space. For example, on the Wikipedia entry Tychonoff plank, the Tychonoff plank is defined to be the space $$(\omega_1+1)\times(\omega+1)$$ and the deleted Tychonoff plank is $$(\omega_1+1)\times(\omega+1)\setminus\{(\omega_1,\omega)\}.$$ These differ from yours in that the vertical dimension reaches only up to $\omega$ rather than $\omega_1$, and so it is a "plank" in being much longer than it is wide. Your space is more like a quarter plane than a plank, and then ultimately your identifications sew copies of it together spiraling around like a long-line version of one of the common Riemann surfaces winding around the origin.

But let me point out that the deleted Tychonoff plank as defined on Wikipedia is not countably compact, since we can take the open set $\omega_1\times(\omega+1)$ together with the horizontal lines $(\omega_1+1)\times\{n\}$. This is a countable cover of the deleted plank, but there is no finite subcover.

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 nested chain of 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.

Terminology. Let me also mention that in my experience what is ordinarily called the Tychonoff plank is a bit different from your space. For example, on the Wikipedia entry Tychonoff plank, the Tychonoff plan is defined to be the space $$(\omega_1+1)\times(\omega+1)$$ and the deleted Tychonoff plank is $$(\omega_1+1)\times(\omega+1)\setminus\{(\omega_1,\omega)\}.$$ These differ from yours in that the vertical dimension reaches only up to $\omega$ rather than $\omega_1$, and so it is a "plank" in being much longer than it is wide. Your space is more like a quarter plane than a plank.

But let me point out that the deleted Tychonoff plank as defined on Wikipedia is not countably compact, since we can take the open set $\omega_1\times(\omega+1)$ together with the horizontal lines $(\omega_1+1)\times\{n\}$. This is a countable cover of the deleted plank, but there is no finite subcover.

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 nested chain of 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.

Terminology. Let me also mention that in my experience what is ordinarily called the Tychonoff plank is a bit different from your space. For example, on the Wikipedia entry Tychonoff plank, the Tychonoff plank is defined to be the space $$(\omega_1+1)\times(\omega+1)$$ and the deleted Tychonoff plank is $$(\omega_1+1)\times(\omega+1)\setminus\{(\omega_1,\omega)\}.$$ These differ from yours in that the vertical dimension reaches only up to $\omega$ rather than $\omega_1$, and so it is a "plank" in being much longer than it is wide. Your space is more like a quarter plane than a plank.

But let me point out that the deleted Tychonoff plank as defined on Wikipedia is not countably compact, since we can take the open set $\omega_1\times(\omega+1)$ together with the horizontal lines $(\omega_1+1)\times\{n\}$. This is a countable cover of the deleted plank, but there is no finite subcover.

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 nested chain of 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.

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 nested chain of 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.

Terminology. Let me also mention that in my experience what is ordinarily called the Tychonoff plank is a bit different from your space. For example, on the Wikipedia entry Tychonoff plank, the Tychonoff plan is defined to be the space $$(\omega_1+1)\times(\omega+1)$$ and the deleted Tychonoff plank is $$(\omega_1+1)\times(\omega+1)\setminus\{(\omega_1,\omega)\}.$$ These differ from yours in that the vertical dimension reaches only up to $\omega$ rather than $\omega_1$, and so it is a "plank" in being much longer than it is wide. Your space is more like a quarter plane than a plank.

But let me point out that the deleted Tychonoff plank as defined on Wikipedia is not countably compact, since we can take the open set $\omega_1\times(\omega+1)$ together with the horizontal lines $(\omega_1+1)\times\{n\}$. This is a countable cover of the deleted plank, but there is no finite subcover.