$\newcommand{\colim}{\operatorname{colim}}$I believe there is a mistake in the argument - as written, the obvious maps $A\to D_n$ are continuous, so the map $A\to\colim D_n$ certainly factors through the maps $D_n\to\colim D_n$. I also don't believe the claim that $\colim D_n$ is indiscrete is accurate - all open subsets of $A$ are also open in that colimit.

Instead, the argument should work if instead you modify $D_n$ so that a subset $U\subseteq A\sqcup\mathbb N$ is open iff $U$ is empty, the whole space or $U$ is *cofinite* and disjoint from $\{0,1,\dots,n-1\}$, with no condition on elemens of $A$.

Now, because $A$ is not discrete, but the subset of $D_n$ on underlying set of $A$ *is* discrete, we see that the maps $A\to D_n$ are not continuous. On the other hand, you can check the colimit in this case is indeed indiscrete - since no cofinite subset of $\mathbb N$ can be disjoint from all $\{0,\dots,n-1\}$, the only two sets which are open in each $D_n$ are the empty set and all of $A\sqcup\mathbb N$. Thus we have a map $A\to\colim D_n$ which doesn't factor through any of the $D_n$.

Finiteness is an easier part - any infinite discrete space $X$ can be written as a colimit of its finite subspaces, $X=\colim X_i$. But $X\to X$ certainly doesn't factor through any of the $X_i$ unless $X$ itself is finite.