$\omega_1 +1$ is compact As we know, $\omega_1$ is not compact. This follows from the fact that $\omega_1$ is a limit ordinal and the open cover of nested clopen intervals will never finitely cover the space.
Now consider $\omega_1+1$. We want to show this space is compact. I’m having trouble showing this, however - primarily because it’s hard to conceptualize how $\omega_1+1$ differs from $\omega_1$.
Any advice on how to show compactness?
 A: The right theorem to prove here is:

Every successor ordinal is compact.

So the important thing isn't the uncountability of $\omega_1$, but rather the "endpoint-having-ness" of $\omega_1+1$ as opposed to $\omega_1$.
This can be proved, appropriately enough, by transfinite induction. Supposing that every $\eta<\alpha$ has $\eta+1$ compact, and let $\mathcal{C}$ be an open cover of $\alpha+1$; we want to show that $\mathcal{C}$ has a finite subcover.
Without loss of generality, assume for simplicity that $\mathcal{C}$ consists only of basic open sets, that is, sets of the form $(\gamma,\eta)$ for $\gamma<\eta\le \alpha+1$. Note that $(\gamma,\eta)=[\gamma+1,\eta)$.
Some $U\in\mathcal{C}$ has $\alpha\in U$; consider $\alpha+1\setminus U$. Since $U$ is a basic open set, we have $$\alpha+1\setminus U=\beta+1$$ for some $\beta<\alpha$. But now applying the induction hypothesis, some finite $\mathcal{D}\subseteq\mathcal{C}$ covers $\beta+1$, and so $\mathcal{D}\cup\{U\}$ is a finite subcover of $\alpha+1$ of $\mathcal{C}$.
