Are there non-homogeneous topological spaces which are "almost homogeneous" in Rudin's sense?

Let $X_n$ be the Cayley graph of the free product $G_n=\frac{1}{2^n}\mathbb{Z}*\frac{1}{2^n}\mathbb{Z}$ (with respect to the two obvious generators). Call edges of $X_n$ that come from the first generator horizontal edges and edges that come from the second generator vertical edges. The inclusion map $G_n\to G_{n+1}$ induces an inclusion map $X_n\to X_{n+1}$. Let $G=\mathbb{Z}[1/2]*\mathbb{Z}[1/2]$ be the colimit of the groups $G_n$, and let $X$ be the colimit of the spaces $X_n$. Note that $G_n$ acts on $X_n$ for each $n$, and these actions glue to give an action of $G$ on $X$. There is also an automorphism of $G_n$ that swaps the two generators which induces an automorphism of $X_n$ that swaps the horizontal and vertical edges, and these glue to give an automorphism $\sigma$ of $X$.

Now I claim $X$ is almost homogeneous. Let $x\in X$ and let $U\subseteq X$ be a nonempty open subset. Then there is some $n$ such that $x\in X_n$ and $U$ contains an open interval of some edge of $X_n$. Applying the automorphism $\sigma$ if necessary, we may assume that $x$ is in a horizontal edge of $X_n$ and that $U$ also contains an open interval of a horizontal edge of $X_n$.

Now note that each edge of $X_n$ gets split in half to become two edges in $X_{n+1}$. So for some $N\geq n$, the open interval that $U$ contains in $X_n$ actually contains an entire horizontal edge of $X_N$. But now the action of $G_N$ is transitive on the horizontal edges of $X_N$, so some element of $G_N$ maps $x$ into the horizontal edge of $X_N$ that $U$ contains. This gives an automorphism of $X$ that maps $x$ into $U$.

On the other hand, $X$ is not homogeneous. To see this, note that if $x\in X$ is a vertex of $X_n$ for some $n$, then $X\setminus \{x\}$ has $4$ connected components (the four "branches" of the tree coming out from the vertex), whereas if $x\in X$ is not a vertex in any $X_n$, then $X\setminus\{x\}$ has only $2$ connected components (the two sides of the tree coming off of the edge that $x$ is in the middle of in each $X_n$).


This example has cardinality $2^{\aleph_0}$; I don't know whether you can get a smaller example but I suspect it is possible to get a countably infinite one. It is easy to see that an almost homogeneous finite space must be homogeneous (consider the minimal nonempty open sets; by almost homogeneity every point must be in one and they all must have the same cardinality, which means the space is a disjoint union of indiscrete spaces of the same cardinality).