Claim: If $(X,\tau)$ is a topological space, $\mathcal B$ a base for $\tau$ and $\mathcal U$ an open cover of $X$ then there is a subcover $\mathcal V \subset \mathcal U$ whose cardinality is not larger than that of $\mathcal B$.
I appreciate some hints, since since is homework. The idea is to remove redundand open sets in $\mathcal U$. My first idea was to remove all $U \in \mathcal U$ which are subeset of the union of other elements of $\mathcal U$, but I failed to show that the remaining subset of $\mathcal U$ still covers $X$.