Another question on 2-connected graphs. These days, I have been trying to solve some questions on $k-$connected graphs. I handed over ones, but still there are ones to do. One of still remaining ones is as follows. I appreciate if someone have a hint or solution.
Assume that $G(V,E)$ is a 2-connected graph of order $n$, and $v_1,\ v_2$ are two vertices in $V$. Let $n_1$ and $n_2$ be two integer positive numbers such that $n_1+n_2=n$. Show that there is a partition $V=V_1\cup V_2$ such that $v_i\in V_i,\ |V_i|=n_i,$ and $G[V_i]$ is connected, for $i=1,2$.
I usually have problem when I need to show the existence of something.
 A: We use induction on $n_1$.
The induction base, $n_1=1$, is obvious: we take $V_1=\{v_1\}$ and $V_2=V-v_1$.
$V_1$ is clearly connected and because $G$ is 2-connected, $V_2$ is connected.
So now assume $n_1>1$ (and $n_1<n-1$).
The induction hypotheses gives us $W_1$ and $W_2$ such that $V=W_1\cup W_2$, $v_1\in W_1, v_2\in W_2$,
$|W_1|=n_1-1$, $|W_2|=n-n_1+1$ and both $G[W_1]$ and $G[W_2]$ connected.
If we find any vertex $u$ in $V_1$ that has some neighbor $w$ in $V_2$ that is not a cut vertex in $W_2$,
we are done (by taking $V_1=W_1+w$ and $V_2=W_2-w$), so we may assume that this situation does not occur.
Specifically $W_2$ cannot be 2-connected and $W_2$ has more than two vertices.
We consider the block-tree decomposition of $W_2$. Let $B$ be a leaf block of this decomposition.
Let $w$ be the (unique) cut vertex belonging to $B$ and $v$ a vertex of $B$ different from $w$ (verify existence!).
Now there must be two internally disjoint paths from $v$ to $v_1$, but our choice of $v$
makes that every path from $v$ to $v_1$ must go through $w$. Contradiction.
