Inductive limit topology and net convergence in a particular example: are the defining spaces open by default?

Question

I'm trying to wrap my head around the concept of inductive limit topologies, especially since it hasn't been that long since I've started learning about category theory.

Let $(X_i)_{i \in I}$ be a family of subsets of a common topological space $E$ with their subspace topology induced by $E$, and suppose that for any given pair $(i,j) \in I^2$, we have $X_i \cup X_j \subseteq X_k$ for some $k \in I$.
We equip $I$ with the following partial order $\leq$: $i \leq j$ iff $X_i \subseteq X_j$, and let for any pair $(i,j)$ such that $i \leq j$ the inclusion map $f_i^j := \operatorname{id}_{X_i \to X_j}$.
$\left((I, \leq), (X_i)_{i \in I}, (f_i^j)_{i \leq j}\right)$ is then a directed system of topological spaces, and $X := \bigcup_i X_i$ (which does not have to be the whole of $E$) given the corresponding final topology, i.e. the topology where a subset $F \subseteq X$ is closed in $X$ iff $F \cap X_i$ is closed in $X_i$ for all $i \in I$, is the inductive limit of said system.

So far so good (I hope). Now, given this setup, let's consider the following question:

Let $(p_a)_{a \in A}$ be a net in $X$ converging to a point $p \in X$.
Is it true that there should exist $i \in I$ such that a subnet of $(p_a)_{a \in A}$ converges to $p$ in $X_i$?

To connect the dots with the second half of the title, what I've first attempted was to show that perhaps the $X_i$s themselves are open sets in $X$, which would make the proof easy: by definition, the convergence of $(p_a)_{a \in A}$ to $p$ implies that for all open neighbourhoods $U$ of $p$, $(p_a)_{a \in A}$ is eventually in $U$, and so in particular this would apply to any $X_i \owns p$. Notice how this would be somewhat overkill.
However, it seemed off to me because in the example I have in mind the $X_i$s would be closed subsets of $E$, and so the $X_i$s are all closed in $X$, as, for any fixed $i \in I$, $X_i \cap X_j$ would be closed in $E$ and so in $X_j$ for all $j \in I$. Evidently, this would prevent $X$ from ever being a connected space if a $X_i \neq X$ were clopen instead of just closed, and so while this does not constitute a true argument, it's counterintuitive enough that I feel I should take another approach...

I also welcome any comments and answers which would instead point out if I've badly translated the problem from its original context, for example maybe it is actually necessary to have $E$ be a locally convex TVS and the $X_i$s be subspaces?
Just like my previous question, this question stems from a statement from the article "Fourier transform for mean periodic functions" by Lázsló Székelyhidi, where the existence of the given $i \in I$ (nonzero compactly supported Borel measure on $\mathbb{R}$ $\mu$) such that the net (of polynomials seen as continuous functions) is in $X_i$ ($= V(\mu) = \{f \in \mathcal{C}(\mathbb{R}) \mid f \ast \mu = 0\}$) comes "[from] the definition of the inductive limit topology".

Feel free to edit and/or re-tag if appropriate or necessary.

Answer 1

For a simple counterexample, let $E=\mathbb{R}$ and take the family of all countable subsets of $\mathbb{R}$ (or even just the countable closed subsets, if you want the $X_i$ to be closed). Then $E$ has the colimit topology, since a subset of $\mathbb{R}$ is closed iff it is sequentially closed. Now let $F$ be the filter of sets that contain a cocountable subset of some neighborhood of $0$, and take any net whose associated filter is $F$. This net converges to $0$, but no subnet contained in any $X_i$ can converge to $0$, since it does not even have any subnets contained in an $X_i$ (the net is eventually outside of every countable set).

As far as I can tell the statement you are asking about in that paper is just an error; the claim may be true in that case but a special argument is needed. Generally, there is very little you can say about convergence of nets in colimits of spaces. See How to characterize the net convergence in final topology? for some more discussion of this.