Question 7.2.16 from S. Morris's Topology without Tears
I'm just trying to get a grip on how these concepts work. So I've written this up for checking.
Prove that $[0,1]$ with the euclidean topology is supercompact:
That is, we need to find a subbasis such that every subbasis cover of $[0,1]$ has a two-element subcover.
We will use the set $\{(-\infty, a): a\in \mathbb{R}\} \bigcup \{(b, \infty):b\in \mathbb{R}\}$ as a subbasis for the euclidean topology on $\mathbb{R}$.
Letting $i \in I$ be elements of an index set, any subbasis cover $\{O_{i}\}$ of $[0,1]$ contains the point $1$ and the point $0$.
Any cover that consists exclusively of $(-\infty, a)$ sets or $(b, \infty)$ sets clearly has one set that contains the entirety of $[0,1]$. Take that one set and union it with any other set in the cover, and those cases are done.
Assume that a given subbasis cover has a mix of $(-\infty, a)$ sets and $(b, \infty)$ sets. WLOG, consider how to include 1.
Considering all subsets $O_i = (-\infty, a_i)$, let $c = \sup_{i\in I}a_i$
Then, either $c \leq 1$ or $c > 1$.
If $c >1$, then $[0,1]\subset O_i = (-\infty, a_i)$ for some $a_i$ such that $1 < a_i < c$.
Take the union of that one set and any other in the cover, and we're done.
If $c \leq 1$, then $c\in O_j = (b_j,\infty)$ for some $b_j$ such that $b_j < c$
Then, there must be some $O_k = (-\infty, a_k)$ such that $b_j < a_k < c$.
So $[0,1] \subset O_j \cup O_k$, and we're done.
Cleaner and fewer cases:
Take the subbase $\mathcal{S}=\{(a,1], a \in [0,1]\} \cup \{[0,b): b \in [0,1]\}$, which is the standard subbase for an ordered space having a maximum $1$ and a minimum $0$ (they're the intersections of $(a,+\infty)$ with $[0,1]$ with $a\in [0,1]$ (if $a$ is not in that range this intersection is either $\emptyset$ or $[0,1]$ hence useless. Likewise for $(-\infty,a) \cap [0,1]$ etc.)).
Let $\{(a_i, 1]: i \in I\} \cup \{[0,b_j): j \in J\}$ be an arbitrary cover by subbasic elements (so all $a_i, b_j \in [0,1]$). $J \neq \emptyset$ as we need to cover $0$ and $I \neq \emptyset$ as we need to cover $1$.
Let $b=\sup\{b_j: j \in J\}$ which is well-defined. Note that $b$ cannot be covered by a set of the form $[0,b_j), j \in J$ or else we would have $b_j > b$ contradicting that $b$ is an upperbound for $\{b_j : j \in J\}$.
So $b$ must be covered by some set $(a_{i_0}, 1]$ with $i_0 \in I$. As $a_{i_0} < b$, $a_{i_0}$ cannot be an upperbound for $\{b_j: j \in J\}$ (because the sup is the smallest upperbound) so there is some $j_0 \in J$ with $a_{i_0}< b_{j_0}$.
It follows that $[0,1] = [0,b_{j_0}) \cup (a_{i_0}, 1]$ and we have a two-element subcover of our subbasic cover.