= Solution
For coordinate differences $a,b$, the elementary inequalities
$$
\max(|a|,|b|)\leq\sqrt{a^2+b^2}\leq\sqrt2\max(|a|,|b|)
$$
give
$$
\boxed{\delta(p,q)\leq d_2(p,q)\leq\sqrt2\,\delta(p,q).}
$$
The identity map between the two versions of the square is bijective, and both it and its inverse are <Lipschitz continuous> maps by these inequalities. It is therefore a <homeomorphism>. In fact the same comparison works on all of $\mathbb R^2$; restriction to the square causes no difficulty.
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