I think this works.

Let $\mathcal{C}$ be the category whose objects are the point of $X$, and define $$ \mathrm{mor}_\mathcal{C}(x,y) = \{ \mbox{closed sets containing both $x$ and $y$} \}. $$ Composition is union.

Now (for example) a sequence $\{ x_n\}$ in $X$ defines a functor $F: \mathbb{N} \to \mathcal{C}$ and a cone from $F$ to $y$ is essentially a single closed set containing the entire sequence and $y$. Since this set must contain the topological limit $x$ of the sequence, this means that the cone factors uniquely through the same closed set viewed as a morphism $x\to y$, so $x$ is the categorical colimit of $F$.

And since the morphism sets are symmetrical, the sequence $\{ x_n\}$ can be viewed as a contravariant functor $G: \mathbb{N}\to \mathcal{C}$, and the topological limit $x$ is the categorical limit of $G$.

I think this doesn't quite work:

Let $\mathcal{C}$ be the category whose objects are the point of $X$, and define $$ \mathrm{mor}_\mathcal{C}(x,y) = \{ \mbox{closed sets containing both $x$ and $y$} \}. $$ Composition is union.

Now (for example) a sequence $\{ x_n\}$ in $X$ defines a functor $F: \mathbb{N} \to \mathcal{C}$ and a cone from $F$ to $y$ is essentially a single closed set containing the entire sequence and $y$. Since this set must contain the topological limit $x$ of the sequence, this means that the cone factors through the same closed set viewed as a morphism $x\to y$, so $x$ is the categorical colimit of $F$.

And since the morphism sets are symmetrical, the sequence $\{ x_n\}$ can be viewed as a contravariant functor $G: \mathbb{N}\to \mathcal{C}$, and the topological limit $x$ is the categorical limit of $G$.

PROBLEM: the factorization is not unique!