Forest graphs $F_1$ and $F_2$ with same vertices but different edges. Prove there is an edge where $F_1$ with it is still a forest I am having quite a hard-time with this question, been thinking about it for a few hours and have not got a clue on how to even start proving this, because it is trivial but proving it has been hard for me.

Given two forests $F_1 = (V,A)$ and $F_2 = (V,B)$ with same vertices group $V$. It is also given that $|B| > |A|$. Prove: there exists an edge $e \in B \backslash A$ where $F_1 \cup \{e\}$ is still a forest.

Any help will be appreciated!
 A: Hint: The fact that $|B|>|A|$ can be used to prove that $F_1$ is not connected. Why does this imply that such an edge exists?
Some more comments: This is an intermediate result; there is an important part of the argument that requires non-connectivity (or something equivalent) to make work.
Notice that in general there is no edge that makes $F_1$ connected. Even if there is, it might not be in $B$. But we are not looking for such an edge, simply one in $B$ which does not produce a cycle in $A$.
A: Add one edges from $B\backslash A$ to $A$ and look if there is a cycle in $A$. If there isn't the problem is solved, otherwise there is exactly one cycle in $A$. In that cycle exists en edge $e$ which is not from $B$, otherwise a cycle will be also in $B$ which is impossible, since $B$ is forest. Remove $e$ from $A$ and continue from beginning. So we can replace all edges in $A$ by edges in $B$.
If there will not be an edge in $B\backslash A$ that will contradict that $|B| > |A|$, since the size of $A$ is not changed.
That means we cant continue infinitely, so there will be an edge in $B\backslash A$ when adding it to $A$ will not form a cycle. 
