I am given sets $A$,$B$ such that there exists $f:A\rightarrow B$ s.t. $f$ is onto $B$.
I am trying to show that $B\le A$
Let $b\in B$, consider $\{a\in A \mid f(a) = b\}$, assuming axiom of choice, this set can be well ordered such that we can pick its smallest element; denote this element by $a_b$.
then $f_0:B\rightarrow A,f_0(b)=a_b$ is an injective function and we are done.
Is this correct?
Your idea is not a good one.
Here's a plausible scenario. Suppose that $S$ is a set such that $S=\bigcup_{n\in\Bbb N}P_n$ where each $P_n$ is a set of cardinality $2$, and there is no choice function from each pair.
It is clear that $S$ can be mapped onto $\Bbb N$, simply by mapping by $f$, $s$ to the unique $n$ such that $s\in P_n$. But now taking $S_n=\{s\in S\mid f(s)=n\}$, it satisfies satisfies $S_n=P_n$. Each $P_n$ can be well-ordered, but it doesn't mean that we can choose, uniformly, from each $P_n$ an element. In this particular case we can't.
The problem is that you need to choose a well-ordering for each $S_n$, and that's something that there is no guarantee that you can do uniformly for each $S_n$.
Instead, given a surjection from $A$ onto $B$, find a family of non-empty sets and define a choice function which is the injection form $B$ into $A$ that you are looking for.