I'm trying to prove that any simple connected graph with at least $3$ vertices ($|V| \ge 3$) has at least $2$ vertices whose removal will not lead to the increment of number of components. In other words, there are at least $2$ vertices that are non-cut vertices.
Attempt at solution:
I tried to prove it using induction by the number of vertices: $n - t \ge 2$, where $n$ is the number of vertices our graph has, and $t$ is a number of cut vertices.
Base
Let $n =3$, then the graph will look like this: $О-О-О$
Clearly, it has one cut vertex (the middle one). So, $3-1 \ge 2$ holds.
Or it can be a full graph, then $3 - 0 \ge 2$ also holds.Assumption
Let $n = k$, and the formula holds, $k - t >= 2$Step
Prove for $n = k + 1$
If we add one vertex to the graph, it could either be a cut vertex or non-cut vertex. In case it is a cut vertex, we have
$\begin{align*}&(k+1) - (t+1) \ge 2\\ &k + 1 - t - t\ge 2\\ &k\ge 2\end{align*}$
In case it is not a cut vertex...and I'm stuck here, because anything can happen. The number of cut vertices might increase, might decrease, or stay the same.
I'm sure there is better and shorter proof to this, and I really hope someone could help me.