My efforts:
Let the uniform space be $(S,\mathcal{U})$.
For a Cauchy net {$x_\alpha$}, the collection of all $B_\gamma$ = {$x_\alpha:\alpha\geq\gamma$}, $\gamma\in I$, is a filter base that extends to a filter $\mathcal{F}$ and an ultrafilter $\Omega$ which converges, say, to $x$, since $S$ is compact. Let $U_x$ be any element of $\mathcal{U}$ such that $\langle x,y\rangle\in U_x$ for some $y$. Then $V_x=\{y:\langle x,y\rangle\in U_x\}$ is a neighborhood of $x$ and thus $V_x\in\Omega$.
Since {$x_\alpha$} is a Cauchy net, for any $U_x\in\mathcal{U}$, there is a $\gamma\in I$ such that $\langle x_\alpha,x_\beta\rangle\in U_x$ whenever $\alpha\geq\gamma$ and $\beta\geq\gamma$, i.e., $B_\gamma\times B_\gamma\subset U_x$. Also $B_\gamma\in\mathcal{F}\subset\Omega$. If $\alpha\geq\beta$, then $B_\alpha\subset B_\beta$.
Then I don't know how to show $B_\gamma\subset V_x$.
We also need to show {$x_\alpha$} converges to the same limit. Assume {$x_\alpha$} also converges to $z$. Let $U_z$ be any element of $\mathcal{U}$ such that $\langle z,y\rangle\in U_z$ for some $y$. Then $V_z=\{y:\langle z,y\rangle\in U_z\}$ is a neighborhood of $z$. There exist $\gamma$ and $\zeta$ such that $B_\gamma\subset V_x$ and $B_\zeta\subset V_z$. Let $\xi=$ max{$\gamma,\zeta$}, then $B_\xi\subset V_x$ and $B_\xi\subset V_z$. It seems no contradiction.
Best Answer
Let $\mathcal F$ be any Cauchy filter on $(S,\mathcal U)$. Since $S$ is compact, $\mathcal F$ has a cluster point $x\in S$. Let $U\in\mathcal U$ be any entourage. Pick an entourage $V\in\mathcal U$ such that $VV\subset U$. Since $\mathcal F$ is a Cauchy filter, there exists a member $F\in\mathcal F$ such that $F\times F\subset V$. Since $x$ is a cluster point of $\mathcal F$, there exists a point $y\in V(x)\cap F$. Then $F\subset V(y)\subset V(V(x))\subset U(x)$. That is $x$ is a limit of $\mathcal F$.