Definitions:
Two paths from $u$ to $v$ are internally disjoint if they have no common internal vertex.
Given $x,y \in V(G)$,a set $ S\subseteq V(G)-\{x,y\}$ is an $x,y-$ separator if $G-S$ has no $x,y$ path.
Let $G$ be a graph on $n\ge 3$ vertices, and let $a,b \in V(G)$ non adjacent vertices.I want to prove the following: the size of every $(a,b)-$ separator is $\ge2$ $\implies$ there exists $2$ internally disjoint $(a,b)-$ paths in $G$.
My approach:
assume that every $(a,b)-$ separator is $\ge2$ and assume by contradiction that there do not exists 2 $(a,b)-$ internally disjoint paths in $G$ . let $G$ be a counter example with $v+e$ minimally. $G$ can't be 2 connected because from whitney's theorem we know that if $G$ is 2 connected there exists $2$ disjoint $(a,b)-$ paths contradiction to our assumption. so $G$ is $1-connected$ so he have a separating set of size $1$. If $a,b$ in different componnents we have an$(a,b)-$separator of size 1 again contradiction so $a,b$ are in the same component - $G'$. Now I'm kinda stuck I'm trying to show that $G'$ is satisfying the conditions and hence we found a smaller graph and that is a contradiction because we assume that $G$ is minimal.
My approach is correct? and if so any ideas how can i continue?
Your approach seems correct, $G^{'}$ is satisfying the requirements as any path from $a$ to $b$ in $G$ must be contained in $G^{'}$ (why?), so every minimal seperator of $(a,b)$ in $G$ must be contained in $G^{'}$.
This implies that there is only $1$ internally disjoint path from $a$ to $b$ in $G^{'}$ and every seperator is of size $\ge 2$.