Let $X,\preceq$ be a partially-ordered set such that every chain $x_n\preceq x_{n+1}\ldots$ has a limit in $\overline X$.
And let $\overline X,\preceq$ be a partially ordered set and a topological space.
But let $\overline X\setminus X$ be a single $\Bbb N$-indexed chain $x'$.
I wanted to apply Zorn's lemma to show that $x'$ is the maximal element.
However, here is where that objective fails: while there is a valid chain $x'$ given by $x'_0,x'_1,\ldots$, and it has a limit in $\overline X$, the idea of every chain having an upper bound breaks down, because:
The chain $x'$, given by $x'_0,x'_1,\ldots$ converges to the limit point $x_0$. Therefore on $\overline X,\preceq$ introducing the concept of a limit point of a chain being an upper bound, reduces $\preceq$ to a preorder. Another way of looking at this, is that the entire chain $x_0\preceq x_1,\preceq x_2\ldots \to x_0$ needs to be in equality.
I don't know how to resolve this problem to show that $x'$ is a maximal element of $\overline X$.
Question
How do I build out the structure further to show $x'$ is either a maximal element, or an element within a larger maximal object?
Attempt
Does creating a further equivalence relation that sets $x_n\sim x_{n+1}$ in each chain, and then let $\preceq$ act on $X/{\sim}$ instead, resolve this problem?
I have another idea, that I take the "identity" chain $x'$ and divide every chain by it to get $\dfrac{x_n}{x'_n}$ to give an isomorphic set of chains which converge to the constant chain $1,1,1\ldots$.
Do either of these look reasonable?
The result is false with only countable limits.
Example: Let $\lambda$ be a limit ordinal with cofinality $>\omega$ (e.g. $\lambda=\omega_1$) and glue the half-ray $\mathbb{R}_{\leq 0}$ to the bottom, identifying $0$. This is our $\bar{X}$. Let $X=\bar{X}-\{-1\}$. Then every countable chain in $X$ either lies completely in $\mathbb{R}_{\leq 0}$ so has a limit there, or eventually in $\lambda$. But $\lambda$ itself is a chain without maximal element, and any countable limit of ordinals below $\lambda$ is $<\lambda$ by supposition.