Graphs such that $|G| \ge 2$ has at least two vertices which are not its cut-vertices Show that every graph $G$, such that $|G| \ge 2$ has at least two vertices which are not its cut-vertices.

Let $P$ be a maximal path in $G$. I claim that the end points of $P$ are not cut vertices. 
Suppose that an end point $v$ of $P$ was a cut vertex. Let $G$ be separated into $G_1,\ G_2,\ \cdots,\ G_k$. It follows that any path from one component to another must pass through $v$ and namely such a path does not end on $v$ and therefore cannot be $P$. Therefore $P$ is contained entirely within some $G_i\cup\{v\}$. But this contradicts the fact that $P$ is maximal for there exists at least one vertex in $G_j$ for $i\neq j$ which connects to $v$ and extends $P$. Therefore $v$ must not be a cut vertex.
