Show that each connected graph that is not a block has at least $2$ blocks, each containing exactly one cut-vertex.

Show that each connected graph that is not a block has at least $$2$$ blocks, each containing exactly one cut-vertex.

I know:

A block of graph $$G$$ is the maximal connected sub-graph of $$G$$ that has no cut-vertices.

So "connected graph that is not a block" means that graph has cut-vertices.

What can I say now?

Given a graph $$G$$, we can associate to it a tree $$T_G$$, sometimes called its block-cut tree, in the following way: the vertex set of $$T_G$$ consists of one vertex for each block of $$G$$, as well as one vertex for each cut-vertex in $$G$$, and for each cut-vertex $$v$$ of $$G$$, we connect the vertex in $$T_G$$ corresponding to $$v$$ to the vertices corresponding to blocks containing $$v$$.
One can show that, as the naming suggests, $$T_G$$ is indeed a tree. A block that contains exactly one cut-vertex corresponds to a vertex of degree one in $$T_G$$; thus, your problem amounts to showing that $$T_G$$ has at least two leaves. This is clearly true under the assumption that $$G$$ itself was not $$2$$-connected (so that $$T_G$$ has at least two vertices).