For future readers, I want to fill in the "relatively easy" gap that the limsup is itself a cluster point of $(x_d)$. For each $d$ let $y_d$ be the sup of the $x_e$'s with $e\geq d$, so that $y_d\rightarrow y$ where $y$ is the limsup. We define a subsequence $(x_{d_n})$ converging to $y$ as follows: inductively define $d_1\leq d_2\leq d_3\cdots$ such that $y_{d_n}-1/n \leq x_{d_{n+1}} \leq y_{d_n}$. Then it is clear that $x_{d_n}\rightarrow y$ as $n\rightarrow \infty$. It's strange that this is achievable with a sequence --- have I made a mistake? — Ehsaan 13 hours ago

@Ehsaan I don't see how would you argue in the proof you suggested that $(x_{d_n})$ is a subset of $(x_d)$. Anyway, to show that $y$ is a cluster point of the given net it is enough to show that if $U$ is a neighborhood of $y$ then the set $\{d\in X; x_d\in U\}$ is cofinal in $D$. — Martin Sleziak 13 hours ago

I see --- the $d_n$'s may not be cofinal. Makes sense. — Ehsaan 13 hours ago

I'm still not convinced. I can see that every cluster point is a limit of a subnet, but that's where you take the definition of "subnet" to simply mean a cofinal function between the indexing sets. If you want an increasing cofinal function as in the wikipedia article you linked, I'm not sure how to do it. (This is the difference between "Kelley" and "Willard" subnets.) — Ehsaan 13 hours ago

Sorry to keep this comment chain going, but I found my problem: a subnet $(x_{f(j)})_{j\in J}$ of a net $(x_i)_{i\in I}$ must have the following cofinality property. For all $i\in I$, there exists $j_0\in J$ such that $f(j)\geq i$ for all $j\geq j_0$. It is not enough to merely have $f(j_0)\geq i$ for a single $j_0$, which is how the wikipedia article defines subnets. With my definition, you can show that a cluster point is the same as a subnet limit. — Ehsaan 8 hours ago

@Ehsaan You can have a look at these posts: Limit superior is a cluster point of a net, Cluster point of a net is a limit of a subnet. In needed, we can discuss this further also in the general topology chatroom. — Martin Sleziak 1 min ago