A connected manifold has a unique dimension $n$, and every point of $X$ then has an open neighbourhood homeomorphic to the open unit ball $\mathbb D^n\subset \mathbb R^n$.
However in the pictured $X$ the points different from $q$ on the hair have an open neigbourhood homeomorphic to $\mathbb D^1$ , whereas the points different from $q$ on the sphere have an open neigbourhood homeomorphic to $\mathbb D^2$.
Since $X$ is connected this proves that it is not a manifold, since it cannot have a unique dimension.
Actually, I think I have a proof of this. I don't even think the smoothness of the homeomorphism is even required. It seems like this is a purely topological result.
I was thinking about this question from your previous post for a while.
Let me know if there are any mistakes.
Theorem. Let $U$ be an open nonempty subset of a smooth $n$-manifold $N$. Let $S$ be a subset of a smooth $m$-manifold $M$. Suppose there is a homeomorphism $f:U\rightarrow S$. Then $m=n$ if and only if $S$ is open in $M$.
Proof. $(\impliedby).$ Suppose $S$ is open in $M$.
Let $p\in S$ (which exists because $U$ is nonempty and $S = f(U)$), let $(X, \phi)$ be a chart of $M$ containing point $p\in S$, and let $(Y, \psi)$ be a chart of $N$ containing point $f^{-1}(p)\in U$.
We claim $f(Y\cap U)\cap (X\cap S)$ is open in $X$. Clearly $Y\cap U$ is an open subset of $U$, so by the homeomorphic property of $f$, the set $f(Y\cap U)$ is an open subset of $S = f(U)$. Since $S$ is open in $M$, the set $f(Y\cap U)$ is open in $M$. Since $S$ is open, $f(Y\cap U)\cap S$ is open in $M$. Therefore, $f(Y\cap U)\cap (X\cap S)$ is open in $X$.
Applying the map $\phi:X\rightarrow\phi(X)$ shows $\phi(f(Y\cap U)\cap (X\cap S))$ is open in $\phi(X)\subseteq \mathbb{R}^{m}$ and hence in $\mathbb{R}^{m}$ (because $\phi(X)$ is open).
By the similar reasoning as in the above two paragraphs, one can show that the set $\psi((Y\cap U)\cap f^{-1}(X\cap S))$ is open in $\mathbb{R}^{n}$ (go over the same reasoning as above; I think you need to do this before you move on to the next paragraph below and I don't think there are shortcuts to this).
Now the composition of homeomorphisms $\psi\circ f^{-1}\circ \phi^{-1}$ sends $\phi(f(Y\cap U)\cap (X\cap S))$ to $\psi((Y\cap U)\cap f^{-1}(X\cap S))$. As we have shown, the former is open in $\mathbb{R}^{m}$ and the latter is open in $\mathbb{R}^{n}$. Also, they are nonempty because the former contains $\phi(p)$ and the latter contains $\psi(f^{-1}(p))$. By applying the last theorem I wrote in this post, we deduce that $m=n$.
$(\implies).$ Suppose $m=n$. Let $p\in S$. Let $(X, \phi)$ be a chart for $M$ containing $p$, and let $(Y, \psi)$ be a chart for $N$ containing $f^{-1}(p)$.
By the homeomorphism property, $f(Y\cap U)$ is open in $S = f(U)$, and by the subspace topology, $f(Y\cap U)\cap (X\cap S)$ is also open in $S$.
By applying $f^{-1}$, the set
$$(Y\cap U)\cap f^{-1}(X\cap S)$$
is open in $U$.
Since $U$ is open, that set is open in $M$; since $Y$ is open, that set is open in $Y$.
By applying $\psi:Y\rightarrow\psi(Y)$, the set $\psi((Y\cap U)\cap f^{-1}(X\cap S))$ is open in $\psi(Y)$, which means it is open in $\mathbb{R}^{n}$.
Then
$$ \phi\circ f\circ\psi^{-1}:\psi((Y\cap U)\cap f^{-1}(X\cap S))\rightarrow \phi(f(Y\cap U)\cap (X\cap S))\subseteq\mathbb{R}^{n} $$
is a continuous injective map, and we know the domain $\psi((Y\cap U)\cap f^{-1}(X\cap S))$ is open in $\mathbb{R}^{n}$ (we can't say the same about the image of the function, however, because not knowing $S$ is open means we don't know whether $X\cap S$ is open in $X$). By applying Theorem 2 of this link, we find that the set
$$ \phi(f(Y\cap U)\cap (X\cap S)) $$
is open in $\mathbb{R}^{n}$. Then $f(Y\cap U)\cap (X\cap S)$ is open in $X$, and hence in $M$ (because $X$ is itself open).
But now we've shown something interesting: for any $p\in S$, there exists an open subset of $M$ of the form $O_{p} = f(Y\cap U)\cap (X\cap S)$ such that
$$ p\in O_{p}\subseteq S. $$
Therefore, we can write $S$ as a union of open sets: $S = \bigcup_{p\in S} O_{p}$ and therefore $S$ is open in $M$.
$$\tag*{$\blacksquare$}$$
Best Answer
Your proof seems to use that $\mathbb{R}^m$ is homeomorphic to $\mathbb{R}^n$ if, and only, if $n=m$, but this statement is highly non-trivial (seems intuitive though) and equivalent to the one for topological manifolds (it is actually the core of the topological invariance of the dimension).
With easy tools (connexity argument), you can prove that $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}$ if, and only, if $n=1$.
As a side note, can you establish with a one-line and really easy proof that $\mathbb{R}^m$ is diffeomorphic to $\mathbb{R}^n$ if, and only, if $n=m$?