= 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, so
$$
|A|+|B|+|\partial_vA|\leq k^2.
$$
The hypotheses make the first two terms greater than $k^2-k$, contradicting $|\partial_vA|\geq k$.
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