A vertex of a minimum vertex cut has a neighbor in every component

I'm trying to understand the solution for the following problem: Prove that ${\kappa}^{\prime}(G)=\kappa (G)$ when G is a simple graph with $\mathrm{\Delta}(G)\le 3$.

The solution goes like this: Let S be the minimum vertex cut, $|S|=\kappa (G)$. Since $\kappa (G)\le {\kappa}^{\prime}(G)$ always, we need only provide an edge cut of size |S|. Let ${H}_{1}$ and ${H}_{2}$ be two components of G-S. Since S is a minimum vertex cut, each $v\in S$ has a neighbor in ${H}_{1}$ and a neighbor in ${H}_{2}$. The solution conntinues from here...I really have no clue why the part "Since S is a minimum vertex cut, each $v\in S$ has a neighbor in ${H}_{1}$ and a neighbor in ${H}_{2}$." is true.

I tried to show it by contradiction but haven't gotten very far: Suppose otherwise; then there exists a vertex $v\in S$ which doesn't have a neighbor in ${H}_{1}$. We can assume that G is connected, hence there is a path joining v with $u\in V({H}_{1})$... and I'm stuck. How do I proceed from here? Or should I try to prove this in a completely different way?

I'm trying to understand the solution for the following problem: Prove that ${\kappa}^{\prime}(G)=\kappa (G)$ when G is a simple graph with $\mathrm{\Delta}(G)\le 3$.

The solution goes like this: Let S be the minimum vertex cut, $|S|=\kappa (G)$. Since $\kappa (G)\le {\kappa}^{\prime}(G)$ always, we need only provide an edge cut of size |S|. Let ${H}_{1}$ and ${H}_{2}$ be two components of G-S. Since S is a minimum vertex cut, each $v\in S$ has a neighbor in ${H}_{1}$ and a neighbor in ${H}_{2}$. The solution conntinues from here...I really have no clue why the part "Since S is a minimum vertex cut, each $v\in S$ has a neighbor in ${H}_{1}$ and a neighbor in ${H}_{2}$." is true.

I tried to show it by contradiction but haven't gotten very far: Suppose otherwise; then there exists a vertex $v\in S$ which doesn't have a neighbor in ${H}_{1}$. We can assume that G is connected, hence there is a path joining v with $u\in V({H}_{1})$... and I'm stuck. How do I proceed from here? Or should I try to prove this in a completely different way?