Is the rational coordinate plane homogeneous?

Question

Let $X$ be a topological space.
For $x, y \in X$ define $x \sim y$, if there is a homeomorphism $f: X \rightarrow X$ such that $f(x) = y$.
It is easy to see that $\sim$ is an equivalence relation on $X$.

$X$ is called homogeneous, if $x \sim y$ for all $x, y \in X$.

Now let $X = (\mathbb R \times \mathbb Q) \cup (\mathbb Q \times \mathbb R)$ as a subspace of $\mathbb R^2$ with the usual topology.

$X$ has been considered in various posts at MSE, for instance here or here. It has some interesting properties:

1. It is easy to see that $X$ is path-connected. Moreover, if $S, T$ are non-empty, open, connected subsets of $\mathbb R$, then $(S \times T) \cap X$ is homeomorphic to $X$. Hence, each point of $X$ has an open neighborhood base of sets homeomorphic to $X$. Thus, $X$ is also locally path-connected.

2. It is easy to see that $X$ is 1-dimensional and of first category.

My question is:

Is $X$ homogeneous ?

Notes

1. Let $(x_1, x_2), (y_1, y_2) \in X$. It is easy to see that $(x_1, x_2) \sim (y_1, y_2)$, if $x_1, x_2, y_1, y_2 \in \mathbb Q$, or, with a little bit more effort, if $x_1$ or $x_2$ is irrational and $y_1$ or $y_2$ is irrational (see addendum). Hence, X has at most the two equivalence classes $[(0,0)]$ and $[(\pi,0)]$.

2. At first glance, it seems $(0,0) \nsim (\pi,0)$, i.e., $X$ is not homogeneous, since $(0,0)$ is the intersection of the two straight lines $\mathbb R \times \{0\}$ and $\{0\} \times \mathbb R$. However, this holds for $(\pi, 0)$ as well: As shown in addendum 2, for every $(x, y) \in X$, $((x, \infty) \times (y, \infty) \cap X) \cup \{(x, y)\}$ is path-connected. Analogous, for the other three quadrants of $(x,y)$. From this, it is easy to find two subsets homeomorphic to $\mathbb R$, whose intersection is $\{(x,y)\}$.

I would also appreciate seeing more references about $X$. In the title I used the ad-hoc name "rational coordinate plane" for $X$. Does anyone know of an established name for it?

Addendum
Here is a proof for note 1:

Let $p, q \in \mathbb Q$. Then $(p,q) \sim (0, 0)$.

PROOF. $\tau: \mathbb R^2 \rightarrow \mathbb R^2, \tau(x,y) = (x-p,y-q)$ is a homeomorphism with $\tau(X) = X$. Hence, $\tau|X: X \rightarrow X$ is a homeomorphism with $\tau(p,q) = (0,0)$.

Let $u, v \in \mathbb R \setminus \mathbb Q, \space p,q \in \mathbb Q$. Then $(u,p) \sim (v, q)$.

PROOF. There are order-isomorphisms $f,g: \mathbb R \rightarrow \mathbb R$ such that $f(\mathbb Q) = g(\mathbb Q) = \mathbb Q$ and $f(u) = v, g(p) = q$. [This is non-trivial: it is based on Cantor's isomorphism theorem. Please let me know, if further elaboration is needed.]
Hence, $f,g$ are also homeomorphisms and $f \times g: \mathbb R^2 \rightarrow \mathbb R^2, (x,y) \mapsto (f(x), g(y))$ is a homeomorphism. It is $(f \times g)(X) = (f \times g)(\mathbb R \times \mathbb Q) \cup (f \times g)(\mathbb Q \times \mathbb R) = (f(\mathbb R) \times g(\mathbb Q)) \cup (f(\mathbb Q) \times g(\mathbb R)) = X$, hence $h = (f \times g)|X: X \rightarrow X$ is a homeomorphism with $h(u, p) = (v, q)$.

Let $(u, p) \in X$. Then $(u,p) \sim (p, u)$.

PROOF. $H: \mathbb R^2 \rightarrow \mathbb R^2, H(x,y) = (y,x)$ is a homeomorphism with $H(X) = X$, hence $H|X: X \rightarrow X$ is a homeomorphism with $H(u,p) = (p,u)$.

Notes to proof

1. All three parts could have been combined into a single proof by a minor modification of part 2. However, I found it clearer this way.
2. All three parts have in common that for appropriate homeomorphisms $f, g: \mathbb R \rightarrow \mathbb R$, the homeomorphism $f \times g$ restricts to a homeomorphism $X \rightarrow X$. This will definitely not work for proving $(\pi, 0) \sim (0,0)$ (if true at all).

Addendum 2
Here is a proof for note 2:

Let $Y$ be a topological space, $x, y \in Y$.
Call a family $(A_n)_{n \in \mathbb N}$ of subsets of $Y$ be a path-provider for $x, y$, if

1. each $A_n$ is path-connected
2. $x \in A_1$
3. for each $n \in \mathbb N$, $\emptyset \neq A_{n+1} \subseteq A_n$
4. for each neighborhood $U$ of $y$, $A_n \subseteq U$ for some $n \in \mathbb N$.

Lemma

If $(A_n)_{n \in \mathbb N}$ is a path-provider for $x, y$, then there is path from $y$ to $x$ in $A_1 \cup \{y\}$, i.e. a continuous function $f: [0,1] \rightarrow A_1 \cup \{y\}$ such that $f(1) = x, f(0) = y$. Hence, $A_1 \cup \{y\}$ is path-connected.

PROOF.
For each $n \in \mathbb N$ choose $x_n \in A_n$, $x_1 = x$ and a continuous map $f_n: [\frac{1}{n+1}, \frac{1}{n}] \rightarrow A_n$ such that $f_n(\frac{1}{n}) = x_n, f_n(\frac{1}{n+1}) = x_{n+1}$.
Then the map $f: (0, 1] \rightarrow A_1, \space f(t) = f_n(t)$, if $\frac{1}{n+1} \le t \le \frac{1}{n}$ is well-defined and continuous, since $([\frac{1}{n+1}, \frac{1}{n}])_{n \in \mathbb N}$ is a locally-finite, closed cover of $(0, 1]$.
Extend $f: [0, 1] \rightarrow A_1 \cup \{y\}$ by $f(0) = y$. By 4., it follows that f is also continuous at $0$. Hence, $f$ is a path from $y$ to $x$.

Now, back to $X$:
For $(x,y) \in X$, define $A_1 = ((x, \infty) \times (y, \infty)) \cap X$ and $A_n = ((x, x + \frac{1}{n}) \times (y, y + \frac{1}{n})) \cap X$ for $n \ge 2$.
$1.$ above implies that $(A_n)_{n \in \mathbb N}$ is a path-provider from $(x+1, y+1)$ to $(x,y)$, hence $((x, \infty) \times (y, \infty) \cap X) \cup \{(x,y)\}$ is path-connected.