Cl(A union B) = Cl(A) union Cl(B) counter example where Cl stands for closure? In my topology book I was asked to prove the above result but instead I think I found a counter example. Consider:

X = {a,b,c,d,e}

A = {a,b}   B = {c,d} and Basis BS = {(a,b,e) , (b,e) , (c,e) , (c,d,e)}, (It's easy to visualize if you draw the basis by writing a,b,e,c,d and making circles left and right accordingly). 

It is a basis because every element is contained in some basis and their is a basis which lies inside intersection of any two basis ( in case of (a,b,e) and (b,e) its (b,e)). 

Now I quote a theorem which says , x belongs to cl(Y) if every basis element containing x intersect Y.

Therefore Cl(A) = A (because some basis of e doesn't intersect A eg (e,c))

Similarly Cl(B) = B, Therefore Cl(A) union Cl(B) = {a,b,c,d}.

Now Cl(AUB) contain e where basis containing e alternatively intersects both A and B, hence Cl(AUB)={a,b,e,c,d}

Could anyone point any mistakes in my argument?
The mistake is that you don’t actually have a base for a topology: both $\{b,e\}$ and $\{c,e\}$ are open, so their intersection, $\{e\}$, must also be open. Thus, $\{e\}$ must also belong to the base. In particular, this means that $e\notin\operatorname{cl}(A\cup B)$.
$
\begin{array}{l}
{{If}\hspace{0.33em}{x}\mathrm{\in}{cl}{\mathrm{(}}{A}{1}\mathrm{\cup}{A}{2}{\mathrm{)}}}\\
{{B}_{\mathit{\epsilon}}{\mathrm{(}}{x}{\mathrm{)}}\mathrm{\cap}{\mathrm{(}}{A}{1}\mathrm{\cup}{A}{2}{\mathrm{)}}\rlap{/}{\mathrm{{=}}}\mathit{\phi}}\\
{{\mathrm{(}}{B}_{\mathit{\epsilon}}{\mathrm{(}}{x}{\mathrm{)}}\mathrm{\cap}{A}{1}{\mathrm{)}}\mathrm{\cup}{\mathrm{(}}{B}_{\mathit{\epsilon}}{\mathrm{(}}{x}{\mathrm{)}}\mathrm{\cap}{A}{2}{\mathrm{)}}\rlap{/}{\mathrm{{=}}}\mathit{\phi}}\\
{{B}_{\mathit{\epsilon}}{\mathrm{(}}{x}{\mathrm{)}}\mathrm{\cap}{A}{1}\rlap{/}{\mathrm{{=}}}\mathit{\phi}\hspace{0.33em}\mathrm{\Rightarrow}{x}\mathrm{\in}{Cl}{\mathrm{(}}{A}{1}{\mathrm{)}}}\\
{{or}\hspace{0.33em}{B}_{\mathit{\epsilon}}{\mathrm{(}}{x}{\mathrm{)}}\mathrm{\cap}{A}{2}\rlap{/}{\mathrm{{=}}}\mathit{\phi}\mathrm{\Rightarrow}{x}\mathrm{\in}{Cl}{\mathrm{(}}{A}{2}{\mathrm{)}}}
\end{array}
$
And the same thing for the another side