Proof of The Squared-Distance Penalization Limit on a Compact Set

Let $\rho:K\times K\to\mathbb{R}$ be the function $\rho(x,y)=d(x,y)$, and regard $\mathbb{R}$ as a metric space through the metric of The Absolute Value Metric on the Real Line.

Continuity of $\rho$. By claim 2 of Continuity of the Projections and of the Distance Function on a Product Metric Space, applied with $T=K\times K$, the function $\rho$ is continuous on $K\times K$ relative to $K\times K$, with respect to the metric $d_{X\times X}$.

Lower semicontinuity of $\psi$. Condition 1 in the definition of a metric gives $0\le\rho(p)$ for every $p\in K\times K$. Hence The Square of a Nonnegative Continuous Real-Valued Function is Continuous, applied in the metric space $(X\times X,d_{X\times X})$ with the subset $K\times K$ and the function $\rho$, shows that $\psi$, which is the function $p\mapsto\rho(p)\,\rho(p)$, is continuous on $K\times K$ relative to $K\times K$. By claim 2 of Semicontinuity Under Negation and Characterization of Continuity, $\psi$ is then lower semicontinuous on $K\times K$.

Nonnegativity. For $(x,y)\in K\times K$ we have $0\le d(x,y)\,d(x,y)$ by Nonnegativity of Squares in an Ordered Field, that is $0\le\psi(x,y)$.

Vanishing exactly on the diagonal. By claim 3 of Zero Products and Elementary Identities in a Field a product of two elements of a field is $0$ only if one of the two factors is $0$, and by claim 1 of that lemma a product with a zero factor is $0$; applied to the two equal factors $d(x,y)$ this gives that $\psi(x,y)=0$ holds if and only if $d(x,y)=0$, which by condition 2 in the definition of a metric holds if and only if $x=y$.

Conclusion. The function $\psi$ therefore satisfies all the hypotheses imposed on the penalty function in Limits of Penalized Maxima on a Compact Set, and the remaining hypotheses of that theorem are those assumed here. Claims 1 to 6 of that theorem consequently hold for $\Phi_{\alpha}(x,y)=u(x)-v(y)-\alpha\,d(x,y)^{2}$.