Union closure for overlapping 2-connected subgraphs

Write $G$ for the finite simple ambient graph and $U=A\cup B$ for the graph specified in the Statement. Every vertex of $U$ lies in $G$, so its vertex set is finite. Each edge of $U$ is an edge of $A$ or $B$, hence is an edge of $G$ with two distinct endpoints in $V(U)$. Thus $U$ is a finite simple subgraph of $G$ containing both $A$ and $B$. In particular, it has at least three vertices.

Choose a common vertex $z$ of $A$ and $B$. For arbitrary $u,v\in V(U)$, choose one of $A,B$ containing $u$ and one containing $v$. Both chosen subgraphs also contain $z$ and are connected, so they supply a path from $u$ to $z$ and a path from $z$ to $v$. Concatenate their vertex lists, listing the joint vertex $z$ only once. Consecutive vertices are adjacent in $U$, so the result is a walk, even if other vertices repeat. The walk-to-path result yields a path from $u$ to $v$ in $U$. Therefore $U$ is connected.

Fix any $x\in V(U)$. For $C=A$ or $C=B$, let $C_x=C-x$ if $x\in V(C)$, and let $C_x=C$ otherwise. The vertices of $A_x\cup B_x$ are exactly the vertices of $U$ other than $x$. An edge belongs to $A_x\cup B_x$ exactly when it belongs to at least one of $A,B$ and is not incident with $x$; these are exactly the edges of $U-x$. Hence $U-x=A_x\cup B_x$.

Each of $A_x,B_x$ is connected: when $x$ belongs to the original subgraph this follows from 2-connectivity, and otherwise that connected subgraph is unchanged. Of the at least two common vertices of $A$ and $B$, at least one is different from $x$ and remains in both $A_x$ and $B_x$. The path-concatenation and walk-to-path argument of step 2.1, through this surviving common vertex, therefore connects every pair of vertices of $A_x\cup B_x$. By step 2.2 this says that $U-x$ is connected. Since $x$ was arbitrary and $U$ is connected with at least three vertices, $U$ is 2-connected.