Assuming the ultrafilter lemma, every net has a universal subnet

Choose an ultrafilter $\mathcal U\supseteq\mathcal F_x$ by [L1]. Let $E=\{(d,A):d\in D,\ A\in\mathcal U,\ x_d\in A\}$, ordered by $(d,A)\preceq(e,B)$ when $d\le e$ and $B\subseteq A$, and put $y_{(d,A)}=x_d$.

The set $E$ is directed. Given $(d,A),(e,B)$, choose $h\ge d,e$. Since $A\cap B$ and the tail $T_h$ belong to $\mathcal U$, their intersection is nonempty; choose an index $k\ge h$ with $x_k\in A\cap B$. Then $(k,A\cap B)$ is above both pairs.

The map $\phi(d,A)=d$ is eventually cofinal: $(d_0,X)$ is an index for every $d_0$, and every later index has first coordinate at least $d_0$. Thus $y$ is a subnet of $x$.

For $S\subseteq X$, [L2] gives $S\in\mathcal U$ or $X\setminus S\in\mathcal U$. In the first case choose any $d_0\in D$. Since $S\cap T_{d_0}\in\mathcal U$, choose $j\ge d_0$ with $x_j\in S$. Then $(j,S)\in E$, and every later value lies in $S$. The complementary case is identical. Thus $y$ is universal.

The constructed $y$ is a universal subnet of $x$.