Vertex cover problem with 2-element vertices
Assuming you want the resulting graph to be disjoint (contain only isolated vertices), then your problem is as hard as the vertex cover problem.  There is a simple reduction.

Let $G^*=(V^*,E^*)$ be an undirected graph, where $V^* = \{v_1,v_2,\dots,v_n\}$.  Define $V = \{v_1,\dots,v_{2n}\}$, and define $E$ by creating edges $((v_i,v_{i+n}),(v_j,v_{j+n}))$ for each $i,j$ such that $(v_i,v_j) \in E^*$.  Note that this automatically satisfies all of your constraints, as there are no pair of edges $(w_1,w_2)$, $(w_1,w_3)$ in $E$ that share a common $v_i$.  Thus, given $G^*$, we can define a new graph $G$ of your form.

Any way to remove vertices in $G$ corresponds immediately to a way to remove vertices in $G^*$, and vice versa.  Moreover, the minimum number of vertices you can remove from $G$ to leave every vertex isolated is equal to the minimum number of vertices you can remove from $G^*$ to leave every vertex isolated.  The latter is exactly the vertex cover problem.  Therefore, we've shown a reduction from vertex cover to your problem: if we could solve your problem for $G$, we could solve vertex cover for $G^*$.