You can also use the theorem that
Let $\mathcal{S}$ be a subbase for $X$. Then $X$ is compact iff every open cover with elements of $\mathcal{S}$ has a finite subcover.
This is Alexander's subbase theorem
and for any ordered space like $[0,1]$ has the following subbase $\mathcal{S}= \{[0,a): a \in [0,1]\} \cup \{(b,1]: b \in [0,1]\}$
Then the compactness of $[0,1]$ follows, using the order completeness:
suppose $\mathcal{O} \subseteq \mathcal{S}$ is an open cover of $[0,1]$.
As $0$ is only covered by a set of the form $[0,a)$ there must be at least one such member in $\mathcal{O}$. So define $A = \{a \in [0,1]: [0,a) \in \mathcal{O}\}$ which is non-empty and bounded above (by $1$) so $a_0 = \sup A$ exists in $[0,1]$. Then $a_0$ is covered by some member of $\mathcal{O}$ and it cannot be of the form $[0,b)$ (as then $b \in A$ and $b >a_0$, so $a_0$ would not be an upperbound of $A$) and so it is covered by some $(b,1] \in \mathcal{O}$. As $b < a_0$ this means that $b$ is not an upperbound for $A$, as $a_0$ is the least upperbound of $A$, so for some $a_1 \in A$ we have $a_1 > b$. But then $[0,a_1)\in \mathcal{O}$ and the two sets $[0,a_1), (b,1]$ from $\mathcal{O}$ clearly cover $[0,1]$.
By the Alexander subbase theorem, $[0,1]$ is compact. In fact, any ordered space with a minimum and maximum which is order-complete is compact (same proof essentially).
The Alexander theorem is proved using AC, so this does not make for a fully "constructive" proof. But it is a useful theorem, e.g. also to show Tychonoff's theorem for compactness of products (product also have a natural subbase), or the compactness of the hyperspace of a compact space in the Vietoris topology.