Thm.
If $G$ is a $k-connected$ graph, and $G'$ is obtained by adding a new vertex $y$, with at least $k$ neighbors in $G$, then $G'$ is also $k-connected$.
Here's what I have so far:
There are 4 cases, but we need only to attend to 3, namely, $y\in S$, the separating set of $G$, $y\notin S,$ but $N(y) \in S$ (Note: $N(y)$ is the neighborhood of $y$), and $y \notin S$ and $N(y) \not\subseteq S$.
Case 1: $y\in S$. If $y \in S$, then we know that $S-\{y\}$ separates G. So $S$ for the new graph $G'$ is at least $k+1$. Thus $S \geq k$.
Case 2: $y \notin S$, but $N(y) \subseteq S$. Since $|N(y)| \geq k$, then $|S| \geq k$.
Case 3: $y \notin S,N(y) \not \subseteq S$. This is where I'm not sure how to proceed. We know that $y$ and $N(y)$ are in a single component of $G'-S$. But then what?
For completeness: Case 4: $y\in S$ and $N(y)\subseteq S$. $S$ for $G'$ is easily at least $k$.
Do cases 1,2,4 look right? And how do I proceed for case 3?
I think that's much easier.
If the number of neighbours of $y$ is $k$, then
In $G'$ you can remove the neighbours of $y$ and disconnect the graph, so the vertex connectivity is less or equal than $k$.
But it can't exist a set $S$ of $k-1$ nodes in $G'$ that disconnect the graph, since $G-S$ remains connected, and also $y$, if not in $S$, is connected to some node, so there's only one connected component.
That means that the vertex connectivity is exactly $k$
If the number of neighbours of $y$ is more than $k$, then the theorem is false:
Take a path with 3 points. It's 1-connected, but if you add $y$ connected to all the other three point, you obtain a graph that is 2-conneccted