Closed real intervals are compact by completeness

Fix a relative open cover $\mathcal{U}$ of $[a,b]$. Let $S$ be the set of $x\in [a,b]$ for which $[a,x]$ has a finite subcover. A member containing $a$ contains a relative neighborhood $[a,a+\delta )$ for some $\delta >0$; choosing $\eta =\min (\delta \mathrm{/}2,(b-a)\mathrm{/}2)$ shows $a+\eta \in S$. Thus $S$ is nonempty and bounded above by $b$. By [A1], $c=\sup S$ exists.

If $c<b$, a cover member containing $c$ contains $(c-\delta ,c+\delta )\cap [a,b]$ for some $\delta >0$. Let $\eta =\min (\delta \mathrm{/}2,(b-c)\mathrm{/}2)>0$. Since $c$ is the supremum, choose $x\in S$ with $x>c-\eta$. A finite cover of $[a,x]$ together with the member covering $c$ covers $[a,c+\eta ]$. This places $c+\eta$ in $S$, contradicting that $c$ is an upper bound. Hence $c=b$.

If $b\in S$, the required finite subcover already exists. Otherwise a cover member containing $b$ contains $(b-\delta ,b]$ for some $\delta >0$. Because $b=\sup S$, choose $x\in S$ with $x>b-\delta$. A finite cover of $[a,x]$ and that member cover all of $[a,b]$.