Recall that feasible solutions satisfying complementary slackness are optimal solutions for the dual problem. So next we show that $p'\vee p''$ satisfies complementary slackness. Suppose $x_{lf}=1$. Then the equations ($*$) hold with equality, and so the conclusion $(**)$ holds with equality as well. Again, a similary argument applies to $p'\wedge p''$. This proves that $(P,\succeq)$ is a lattice.

Finally we show that $(P\succceq)$ is complete. Let $A\subset P$ be a set of payoffs. The $\sup\{p:p\in A\}$ is $\bar p$ such that $\bar w_l=\sup\{w_l:p\in A\}$ and $\bar\pi_f=\inf\{ \pi_f:p\in A\}$. For $(P,\succeq)$ to be complete, $\bar p\in P$. That is, $\bar p$ has to be an optimal solution to the dual.

We need to show that $\bar p$ satisfies feasibility and complementary slackness. Choose a matching $x$. For all $\epsilon>0$ there is a payoff in $A$ with $w_l^\epsilon\leq \bar w_l< w_l^\epsilon+\epsilon$ and $\pi_f^\epsilon\geq\bar\pi>\pi_f^\epsilon-\epsilon$. Then \begin{equation*} v_{lf}-\epsilon\leq w_l^\epsilon+\pi_f^\epsilon-\epsilon\leq \bar w_l+\bar \pi_f. \end{equation*} Letting $\epsilon\to0$ and taking the $\limsup$, $\bar w_l+\bar\pi_f\geq v_{lf}$, $\bar p$ is seen to be feasible. If $x_{lf}=1$ then \begin{equation*} \bar w_l+\bar\pi^\epsilon_f\leq w^\epsilon_l+\epsilon+\pi^\epsilon_f=v_{lf}+\epsilon \end{equation*} Let $\epsilon\to 0$ and take the $\limsup$ to see that $\bar w_l+\bar\pi_f\leq v_{lf}$. This establishes equality, which demonstrates complementary slackness and thereby proves the theorem. ∎