From Schaum's Topology Outline, Chapter 7, Supplementary problem #39.
Prove: Let $f:(X,\mathscr{T})\rightarrow (Y, \mathscr{T*})$ be open and onto, and let $\mathscr{B}$ be a base for $\mathscr{T}$. Then {$f [B]:B\in \mathscr{B}$} is a base for $\mathscr{T*}$.
I think this cannot be proved in general. [Edit: The book does not specify whether or not $f$ is continuous. Other problems in the same section do occasionally say "Let $f$ be continuous", so I don't think we can assume continuity here.]
Here is what I believe is a counterexample. Please let me know if I'm missing something:
Let $X = \{a,b,c,d\}$ and $\mathscr{T} = \{X, \{a,b\}, \{c,d\}, \emptyset\}$
Let $ Y = \{1,2,3\}$ and $\mathscr{T*} = \{Y, \{1,2\}, \{2,3\}, \{2\}, \emptyset\}$
Define the function, $f:X\rightarrow Y, f(a) = 1; f(b) = f(c) = 2; f(d) = 3$
Then, $f$ is open because it maps open sets to open sets, and $f$ is onto because every element of $Y$ is mapped to.
Now, let $\mathscr{B} = \{\{a,b\},\{c,d\}\}$ be a base for $\mathscr{T}$.
This is base because its elements are open sets and every open set in $\mathscr{T}$ is a union of members of $\mathscr{B}$. (NB: the empty union provides the empty set)
But then $\{f [B]:B\in \mathscr{B}\}=\{\{1,2\},\{2,3\}\}$, which is NOT a base for $\mathscr{T*}$ because you cannot obtain the open set $\{2\}$ with unions of elements of this set.
So the general statement of the proof is false.
Speculation: Perhaps the problem should read "into" instead of "onto"?