Condition (i) implies that $u\circ p_{\mathcal{X}}=u\circ i_{\mathcal{X}}\circ p_{\mathcal{X}}$ is continuous for all $\mathcal{X}$, so that $v:=\inf_{\mathcal{X}}u\circ p_{\mathcal{X}}$ is usc. It follows that $v\geq u^{*}$, since $v\geq u$ by (ii). Conversely, for each $x\in X$ we have $\lim_{\mathcal{X}}p\mathcal{X}(x)=x$, hence

$$u^{*}(x)\geq\limsup_{\mathcal{X}}u^{*}\circ p_{\mathcal{X}}(x)\geq\inf_{\mathcal{X}}u\circ p_{\mathcal{X}}(x)=v(x),$$

which shows that $v=u^{*}$. Finally, (ii) shows that $\mathcal{X}^{\prime}\geq\mathcal{X}\Rightarrow u\circ p_{\mathcal{X}^{\prime}}\leq u\circ p_{\mathcal{X}}$, hence $v\circ p_{\mathcal{X}}=u\circ p_{\mathcal{X}}$, which is equivalent to the last assertion. ∎