I need some help with the following problem:

Let $\displaystyle f: (X,T) \to (Y,T*)$ be an open and onto function from the topological space (X,T) onto (Y,T*) and let $\displaystyle \beta$ be a base for T. Prove that {$\displaystyle f[ B ]: B \in \beta$} is a base for T*.

I understand that the image if any open set is open and that the image of an open set is the union of the images of the base elements of T. But since there is no continuity requirement Y can have an open set that has as an inverse image a set that is neither open nor closed and so isn't the union of the images of the bases elements of T.

I think in fact the following trivial example shows this can't be true (which means I must be not understanding something fundamental):

X=Y={a,b,c}

Let T be the indiscrete topology on X, so that the only open sets are X and $\displaystyle \emptyset$. Then the only base for X is X itself.

Define f: f(x)=x, but let T* be a different topology: say the open sets are X, $\displaystyle \emptyset$ and {a}.

f is still open an onto, but the image of all the base elements of T does not form a base for T*, since it misses {a}, which has the set {a} in X as an inverse image, which is neither open nor closed in (X,T).

Any help telling me why my example is wrong and how to prove this would be appreciated.