If $B$ is a subbasis for $T$, $B'$ for $T'$, $C:= B \times B'$ is closed under intersections If $(X,T), (X',T')$ are topological spaces and $B$ is a subbasis for $T$ and $B'$ is a subbasis for T', then $C:= B \times B'$ is closed under intersections. This is stated without proof in my notes.
So... I would take two arbitrary elements in $C$ say $(X_1, X'_1)$ and 
$(X_2, X'_2)$ and I would have to show that their intersection is in $C$. My question is what if $X_1 \cap X_2$ is not an element in $B$ ? Would this not contradict that $C$ is closed under finite intersections?
Edit: This is part of a theorem : 
Suppose $(X, T )$ and $(X', T')$ are topological spaces. If $B$ is a subbasis for $T$
and $B'$ is a subbasis for $T'$, then $C := B×B'$ is a basis for the product topology on $X \times X'$.

No, not even if B and B' are bases.
Let B = {{0}} be a base for {0} and
B' = { {n}, {n,m,q} : distinct n,m,q in N }
be a base for the discrete topology of N.  
Of note is that {1,2,3} $\cap$ {2,3,4} is not in B'.
Perhaps if B and B' were minimal bases,
then B×B' would be closed under intersection. 
Alternatively, if B and B' are topologies,
then B×B' is intersection closed.
