Set of functions that form a basis for a Topology I'm trying to prove the following statement: Define $X=\lbrace f| f: [0,1]\rightarrow[0,1]\text{ is a function}\rbrace$. For each subset $A$ of $[0,1]$, define
\begin{equation}
B_A=\lbrace f\in X|f(x)=0, \forall x \in A \rbrace.
\end{equation}
Prove that the set $B=\lbrace B_A|A \subseteq [0,1]\rbrace$ is a basis for a topology on $X$.
I tried doing the following: let $f \in X$ with $f \in B_{A_1} \cap B_{A_2}$. Then I need to prove that there is a basis element $B_A$ in $B$ such that $f \in B_A \subseteq B_{A_1} \cap B_{A_2}$. I got stucked in here: if $f \in B_{A_1} \cap B_{A_2}$ then
\begin{equation}
f(x)=0, \forall x\in B_{A_1} \cap B_{A_2}.
\end{equation}
How can I proceed from here? Only hints, not answers, please.

If a map $f$ vanishes on $A_1$ and $A_2$, on which minimal subset (for the inclusion order) does $f$ vanishes?
Note: you also have to prove that the base elements cover $X$.
