 Solution

This is true. A simplicial initial segment in $[k{]}^2$ having both it and its complement larger than $(k^2-k)/2$ has an external vertex boundary of at least $k$. By the vertex-isoperimetric inequality, the same is true for every such $A$. If $A$ and $B$ were disjoint with no edge between them, then $B$ would avoid both $A$ and its external boundary, soThe hypotheses make the first two terms greater than $k^2-k$, contradicting $|{\partial }_{v}A|\ge k$.