Closed Intervals of $\omega_1$ Are Compact

If you want to use a more general approach: for ordered spaces, compactness = order completeness (every subset has a supremum and an infimum), and for closed intervals of $\omega_1$ (where the induced topology = the order topology) this follows easily: we have a min for every non-empty subset, so certainly an infimum, and being bounded above gives us a minimum for the set of upper bounds hence a supremum as well, which by being a closed interval must be in the closed interval too.

OUTLINE: Without loss of generality start with a cover of $(\alpha,\beta]$ by intervals of the form $(\xi,\eta]$. There must be one of the form $(\xi_0,\beta]$. Then there must be one of the form $(\xi_1,\xi_0]$. In this way construct a strictly decreasing sequence $\langle \xi_n:n\in\omega\rangle$. Therefore?

First note that $[1,\omega^2]$ is in fact isomorphic to $\omega^2+1$ as an ordinal, since the closed interval implies that $\omega^2$ is in the interval, and it is a maximal element.

Secondly, note that if $[\alpha,\beta]$ is a closed interval of ordinals then it is isomorphic to an ordinal via a continuous and open order preserving isomorphism, therefore it is homeomorphic to an ordinal.

By the above claim we have that $[\alpha,\beta]$ is isomorphic to a successor ordinal (since the set is isomorphic to a successor ordinal: $\mathrm{otp}(\beta-\alpha)+1$) and successor ordinals are always compact in the order topology since a cover of $\gamma+1$ can be made into a decreasing chain of ordinals - which is finite.

Claim: Suppose $[\alpha,\beta]$ is a closed interval of ordinals, then it is homeomorphic to a successor ordinal $\gamma$.

Proof: Since as an ordered set $[\alpha,\beta]$ is well-ordered there exists a unique ordinal and a unique isomorphism $f$ between the interval and the ordinal. First it is obvious that $\gamma$ is a successor ordinal, since $\beta$ is the maximal element in $[\alpha,\beta]$ therefore the order type of $\gamma$ has a maximal element, which can only occur at successor ordinals.

Continuity in well-ordered topologies is to say that if $\delta$ is a limit ordinal then $f(\delta)=\sup\{f(\tau)\mid\tau<\delta\}$. We recall that $f$ is such that $f(\alpha)=0; f(\tau+1)=f(\tau)+1;$ and $f(\delta)=\sup\{f(\tau)\mid\tau<\delta\}$ for limit ordinals $\delta$, so indeed the mapping is continuous.

Lastly we need to show that it is indeed open, which is to say that the image of an open interval is open. This is another result of the definition of $f$, since if $(\alpha+\epsilon,\beta')=[\alpha+\epsilon+1,\beta')$ is an open interval in $[\alpha,\beta]$ then it holds that $f''[\alpha+\epsilon+1,\beta')=[f(\epsilon)+1,f(\beta')]$.

Therefore we have that indeed $[\alpha,\beta]$ is homeomorphic to $\gamma+1$ for some ordinal $\gamma$, and therefore compact.$\square$

Note that if $\gamma$ is countable then $\gamma+1$ is a Polish space, that is a separable, completely metrizable metric space. However compactness does not require metrizability in the case of ordinals, and indeed any successor ordinal is a compact space (when equipped with the order topology).