Simple Failings of Axiom of Choice in Top
The statement "every surjective function has a right inverse" is equivalent to AC. The corresponding statement in $\mathsf{Top}$ would be "every continuous surjection has a continuous right inverse", which is false (for instance, pick your favourite continuous bijection that isn't open).
This is mentioned in the link you posted, though, so I'm not totally sure if it's the kind of thing you're looking for.
Let $F$ be a field object in the category $\mathbf{Set}$. Every $F$-vector-space object in $\mathbf{Set}$ is isomorphic to a direct sum of copies of $F$. This is equivalent to the fact that every vector space has a basis, which requires the Axiom of Choice.
We consider the following generalization: let $F$ be a field object in the category $\mathbf{Top}$. Is every $F$-vector-space object in $\mathbf{Top}$ isomorphic to a direct sum of copies of $F$?
Let $F= \mathbb{R}$ equipped with the usual Euclidean topology. Take $1 \leq p < q$. Assume for a contradiction that $\ell^p$ and $\ell^q$ are both direct sums of copies of $\mathbb{R}$. Both spaces have cardinality $|\mathbb{R}|$, and have bases of that same cardinality. Since all bases of a vector space are equipotent, this would mean that they're both isomorphic to the direct sum of $|\mathbb{R}|$-many copies of $\mathbb{R}$. But by Pitt's theorem $\ell^p$ and $\ell^q$ are not linearly homeomorphic, a contradiction.
We should note that the counterexample given above is honest in the sense that Zermelo-Fraenkel set theory without Choice does not prove the existence of a basis for $\ell^p$. Dishonest counterexamples abound (and their existence should not surprise us, given that the failure of Choice is by far not the only property distinguishing $\mathbf{Set}$ from $\mathbf{Top}$).
Let $S$ be a non-initial object in the category $\mathbf{Set}$. Then we can find arrows $e: 1 \rightarrow S$, $i: S \rightarrow S$ and $m: S \times S \rightarrow S$ making $S$ into a group object. Given Choice, this follows immediately from the existence of cyclic groups (for the finite case) and the upward Löwenheim–Skolem theorem (for the infinite case). Moreover, there's an easy argument showing that this result strictly requires Choice.
Again, the corresponding statement fails for objects $S$ in $\mathbf{Top}$, even if we restrict our attention to very nice spaces. For example, the bouquet of $n$ circles does not admit a topological group structure for any $n \geq 2$.
Now take your favorite non-trivial group $G$, and consider the category $G$-$\mathbf{Set}$. An object of this category is a set equipped with a left-action of $G$, and a morphism is just an equivariant map between two such sets.
Unlike $\mathbf{Top}$, $G$-$\mathbf{Set}$ has very similar properties to $\mathbf{Set}$: it forms what is known as a Boolean topos. However, the analogue of the statement above fails even for objects of the category $G$-$\mathbf{Set}$. For consider the underlying set of the group $G$ equipped with the left multiplication action of $G$, and denote it $G^\star$. To make $G^\star$ into a group object, one has to construct a morphism from the terminal object $e: 1 \rightarrow G^\star$. The terminal object $1$ of $G$-$\mathbf{Set}$ is the one-element set $\{0\}$ equipped with the trivial $G$-action, so one would have $\forall g \in G. e(0) = e(g \cdot 0) = g e(0)$. Therefore, if $G$ is non-trivial, no suitable $e$ exists.
This relates directly to the failure of the Axiom of Choice in $G$-$\mathbf{Set}$: the argument above proves that the unique map $G^\star \rightarrow 1$ does not have a section.