Graph with 3 blocks which any two of them are connected.

For a graph with 3 blocks like $B_1, B_2, B_3$ if every two of them are common in a vertex like $v,u,w$ then non of these vertices are cut-vertex. Because $G-\{v\}$ is still a connected graph. Am I right?

Let $B_1$, $B_2$ and $B_3$ be distinct pairwise-intersecting blocks.

Let $u \in B_1 \cap B_2$, $v \in B_2 \cap B_3$ and $w \in B_1 \cap B_3$. Let $P_1$ be a path in $B_1$ from $w$ to $u$, $P_2$ be a path in $B_2$ from $u$ to $v$ and $P_3$ be a path in $B_3$ from $v$ to $w$. These three paths are edge-disjoint by $(a)$ and node-disjoint (that is they do not have common inner nodes) by $(b)$. Thus if $u$, $v$ and $w$ are distinct, the concatenation of $P_1$, $P_2$, $P_3$ is a circuit containing $u$, $v$ and $w$, from which we get that $w \in B_2$, contradicting $(b)$. Hence $u = v = w$.

So it is not that $u$, $v$ and $w$ are not cut-vertices, but instead that they are the same cut-vertex.