Solution:
Disregard all edges except those used in the path from $A$ to $B$, and for each of those let the multiplicity be the number of times it was traversed. Let the degree of a vertex be the sum of the multiplicities of its edges. The key is to notice that the degree of every vertex except $A$ and $B$ must be even. For as we traverse the path from $A$ to $B$ we increase the degree by $2$ each time we pass through a vertex. But at the start of the path, as we leave $A$, we only increase its degree by $1$. Similarly as we arrive at $B$ for the last time.

Now construct a path from $B$ as follows. Since $B$ has odd degree it must have an edge of odd multiplicity. Suppose the edge connects $B$ to $C$. Follow that edge and reduce its multiplicity by one, so that $B$'s degree and $C$'s degree are each reduced by one. Now $C$ has odd degree, so it must have an edge of odd multiplicity. Repeat. Since there are only finitely many edges we must eventually be unable to continue the path. But the only way that can happen is if we reach $A$.