In a Hausdorff topological space, let net $(x_d)_{d \in D}$ converge to $x$. The set $$ (\cup_{d \in D} \{x_d\}) \cup \{x\} $$ consisting of its terms and the limit need not be compact: in $\mathbb{R}$ with its usual topology, the net of rationals in $[0,1]$ with the standard order converges to $1$, yet $[0,1] \cap \mathbb{Q}$ is not compact.
But is there always a subnet of $(x_d)_{d \in D}$ whose terms, together with limit $x$, form a compact set?
I came across this claim in a recent paper, but don't see how to prove this.
Let's try this: $X = [0,\omega_1]$ a set of ordinals in the order topology, where $\omega_1$ is the least uncountable ordinal. Let $D \subseteq X$ be all of the non-limit ordinals. It is directed in its usual order, the net $x_d = d$ converges to $\omega_1$. Now try to find your subnet.