The problem states: Let $X$ be a completely regular space such that $|X|<c=|\mathbb{R}|$. Prove that there exists a basis $\mathcal{B}$ for the topology on $X$ such that for all $B\in\mathcal{B}$, $B$ is both open and closed.
I let $\mathcal{B}$ be the collection of all clopen sets since any other choice for $\mathcal{B}$ would yield a coarser topology. It is easy to verify that $\mathcal{B}$ forms a valid basis. Let $\tau$ be the topology on $X$ and let $\tau_\mathcal{B}$ be the topology generated by $\mathcal{B}$. Then clearly $\tau_\mathcal{B}\subset\tau$. I couldn't figure out how to show that $\tau_\mathcal{B}=\tau$, so I assumed it wasn't and tried coming to a contradiction. If $\tau_\mathcal{B}\neq\tau$, then that means there must exist an open set $U\in\tau$ and a point $x\in U$ such that every clopen set containing $x$ must intersect $X\setminus U$. I thought that maybe using this along with the completely regular condition, I could prove the existence of some surjective function $f:X\to[0,1]$, thus contradicting the cardinality restriction, but I couldn't find any way of doing that. Any ideas?
As you say, let us take all Clopen sets in our topology. To show that it is a base, we need to show that for every open set $O$ and $p \in O$ there is a member of the base $M$ with $p \in M \subseteq O$. Now how can we find $M$? Well because the space is completely regular, we have a function $f$ with the image of $p$ being $0$ and the image of $O^{c}$ (a closed set that does not contain $p$) being $1$. Now what is the inverse image of $1$? It will be a superset of $O^{c}$. It will also be closed (as the inverse image of a closed set via a continuous function) and so its complement will be open, will contain $p$ (because $p$ maps to $0$) and will be a subset of $O$. Now can we see that this set if also closed? We must use the cardinality condition for this. OK so let $I$ be the image of $X$ under $f$. This cannot contain any open intervals (because if it did then $X$ would have cardinality greater than or equal to $|R|$. Hence every point of $I$ is open and closed in the subspace topology (of $I$ derived from $R$) and so the inverse image of $1$ must be open and closed, its complement must be open and closed and the set of Clopen sets is a basis for $X$.