Compactness of second-countability of $\omega$X$\omega_1$

Question

Please discuss the following properties of the product space consisting of $\omega$X$\omega_1$:

1. Is it compact?
2. Is it 2nd countable?

$\omega$ is the first infinite ordinal and $\omega_1$ is the first uncountable ordinal. The topology is the product topology.

I know that the topological space $[0,\omega_1)$ is not compact, leading me to conclude that the answer for (1) is 'no' using the cartesian product definition of two compact spaces, but here (at least) one of them fails to be compact. Anything pointing me in the right direction will be helpful.

Answer 1

HINTS: As Arthur Fischer said in the comments, the deleted Tikhonov plank is not the space that you described, but rather the space obtained by deleting the point $\{\langle\omega,\omega_1\rangle$ from the product space $[0,\omega]\times[0,\omega_1]$.

1. Show that the deleted Tikhonov plank is a non-closed subset of a compact Hausdorff space.
2. The deleted Tikhonov plank contains a subspace homeomorphic to $[0,\omega_1]$.

Added: It turns out that the space in question is actually $\omega\times\omega_1$, so you can simply use the fact that the space contains a clopen subspace homeomorphic to $\omega_1$ to answer both questions.