Problem: Let $\mathcal{S}$ be a subbasis for a topology $\mathcal{T}$ on the set $\mathbb{R}$. If all of the closed intervals $[a,b]$, with $a<b$, are in $\mathcal{S}$, prove that $\mathcal{T}$ is the discrete topology.
Solution: In order to show that $\mathcal{T}$ is the discrete topology, it's enough to show that every subset is (closed) open. Let $B\subseteq X$. Notice
$$
B=\bigcup_{b\in B}\{b\}.
$$
Consider two closed intervals $[a,b]$, with $a<b$, and $[b,c]$, with $b<c$. These are elements of the subbasis, thus there intersection $\{b\}=[a,b]\cap[b,c]$ is an element of the basis. Therefore, for any $b\in\mathbb{R}$ the singleton $\{b\}$ is open. As such, $B$ being the union of open sets is itself an open set.
Best Answer
Yes, that proof works. Note that the clause $a < b$ stops one to use $[a,a]=\{a\}$ as a closed interval (though it is a closed interval in any ordered set really, only a degenerate one). But any point $x$ in $\Bbb R$ has a point below it, like $x-1$ and one above it, say $x+1$, so $\{x\}= [x-1,x] \cap [x,x+1]$ is in the base for any $x$. That last argument is a bit more precise than yours IMHO, as it starts out from an arbitrary $x$. If we'd use the same subbase in $[0,1]$, say, $\{0\}$ and $\{1\}$ would not have been open and the topology not discrete.