We are given the collection, $$\mathcal{B}=\{[a,b]\mid a,b\in\mathbb{R}\}.$$ We want to show that this collection is a basis for a topology on $\mathbb{R}$. To show this we show the following:
First, for all $x\in\mathbb{R}$ there exists an element in $\mathcal{B}$, namely $[x-\epsilon,x+\epsilon]$, with $\epsilon\in\mathbb{R}$ which contains $x$.
Second, let $x\in[a,b]\cap[c,d]$. We want to show there exists a third element, say $B$ in $\mathcal{B}$, such that $x\in B\subset[a,b]\cap[c,d]$. Consider, $B=[c,b]$.
So we have shown that $\mathcal{B}$ is a basis for some topology on $\mathbb{R}$.
The question then asks me to show that the topology generated by this basis is the discrete topology. This amounts to showing that the basis elements of either topology is open in the other.
The discrete topology is generated by single element sets. So, we have trivially that all basis elements in $\mathcal{B}$ are open in the discrete topology. Similarly, we consider the basis element $[a,a]\in\mathcal{B}$. We have that for a given $\{a\}\in\mathcal{C}$, where $\mathcal{C}$ is the discrete basis, there exists a $[a,a]\in\mathcal{B}$ such that all the basis elements in $\mathcal{C}$ are open in the topology generated by $\mathcal{B}$.
So, I conclude that the topologies are equal. My question is the following. Is it okay for me to treat the element $[a,a]=\{a\}$ as an element of $\mathcal{B}$ in this proof? The question imposes no restriction on the value of $a$ or $b$, so I think it's okay. However, I'm not sure.
Yes, that's totally fine. If you want to avoid it, you could also write $$ \{a \} = [a-1,a] \cap [a, a+1]. $$