of $T_X$, and $U_Y$ the set of vertices only appearing in bags of $T_Y$. Then $U_X, X \cap Y, U_Y$ is a partition of $V(G)$ such that no edge has an endpoint in $U_X$ and $U_Y$. Each component of $G-(X \cap Y)$ is contained entirely within $U_X$ or $U_Y$. Say $Q \subseteq V(G)$ is *large* if $|Q \cap S| > \frac{1}{2}|S-(X \cap Y)|$.

If neither $U_X$ or $U_Y$ is large, then no component of $G-(X \cap Y)$ is large. Hence $X \cap Y$ is a $(|X \cap Y|, S,\frac{1}{2})$-separator. Since $|X \cap Y| \leq |Y| \leq k$, this is sufficient.

Alternatively, for all edges $XY \in E(T)$, exactly one of $U_X$ and $U_Y$ is large. (If both sets are large, then $|S-(X \cap Y)| = |U_X \cap S| + |U_Y \cap S| > |S-(X \cap Y)|$, which is a contradiction.) Orient the edge $XY \in E(T)$ towards $X$ if $U_X$ is large, or towards $Y$ if $U_Y$ is large.

Now there must be a bag $B$ with outdegree 0. If $B$ is a $(|B|, S,\frac{1}{2})$-separator, then as $|B|=k$, the result is achieved. Otherwise, exactly one component $C$ of $G-B$ is large. The vertices of $C$ only appear in the bags 