Solution:
Take any path from $A$ to $B$. Suppose it is $A = A_0$, $A_1$, ..., $A_n = B$. We show by induction on $r$ that we can find two disjoint paths from $A$ to $A_r$.

If $r = 1$, then take any vertex $C$ distinct from $A$ and $A_1$. Take any path from $A_1$ to $C$ which does not go through $A$. Now take any path from $C$ to $A$ which does not go through $A_1$. Joining these two paths together gives a path $p$ from $A$ to $A_1$ which does not involve the edge $AA_1$. Then $p$ and the edge $AA_1$ are the required disjoint paths.

Suppose now we have two disjoint paths $A$, $B_1$, $B_2$, ..., $B_s$, $A_r$ and $A$, $B_t$, $B_{t - 1}$, ..., $B_{s + 1}$, $A_r$ and we wish to find two disjoint paths joining $A$ and $A_{r + 1}$. Take a path between $A$ and $A_{r + 1}$ which does not include $A_r$. If it also avoids all of $B_1$, ..., $B_t$, then we are home, because it is disjoint from the alternative path $A$, $B_1$, $B_2$, ..., $B_s$, $A_r$, $A_{r + 1}$. If not, let $B_i$ be the first of the $B$'s on the path as we move from $A_{r + 1}$ to $A$. This allows us to construct two disjoint paths from $A$ to $A_{r + 1}$. One path goes from $A$ to $B_i$ and then from $B_i$ to $A_{r + 1}$. The other path goes around the other way to $A_r$ and then along the edge to $A_{r + 1}$.

[Explicitly, if $i \leq s$, then the paths are $A$, $B_1$, $B_2$, ..., $B_i$, ..., (new path) ... $A_{r + 1}$ and $A$, $B_t$, $B_{t - 1}$, ..., $A_r$, $A_{r + 1}$. If $i > s$, then the paths are $A$, $B_t$, $B_{t - 1}$, ..., $B_i$, ..., (new path) ... $A_{r + 1}$ and $A$, $B_1$, ..., $B_s$, $A_r$, $A_{r + 1}$.]

Hence, by induction, there are two disjoint paths from $A$ to $B$.