Is $D=\{(x,0) \in \mathbb{R}^2: x \in \mathbb{R}\} \cup \{(0,y) \in \mathbb{R}^2: y \in \mathbb{R} \} $ a 1d or 2d manifold

Question

Suppose you have the set $D=\{(x,0) \in \mathbb{R}^2: x \in \mathbb{R}\} \cup \{(0,y) \in \mathbb{R}^2: y \in \mathbb{R} \} $ with the subspace topology of $\mathbb{R}^2$.
Is D a manifold?
My understanding is that for the set D to not be a manifold i have to show that there does not exist a homeomorphism $f:U \subset D \rightarrow f(U) \subset \mathbb{R}^2$ where $f(U)$ open in $\mathbb{R}^2$.

My intuition says that it is not a manifold, but what confuses me is that it is a subspace of $\mathbb{R}^2$ with the subspace topology.

Edit: Can it be a 1d manifold with charts:

$f_1:(0,+\infty) \rightarrow \mathbb{R}^2$ with $f_1(x)=(x,0)$
$f_2:(0,+\infty) \rightarrow \mathbb{R}^2$ with $f_2(x)=(-x,0)$
$f_3:(0,+\infty) \rightarrow \mathbb{R}^2$ with $f_3(y)=(0,y)$
$f_4:(0,+\infty) \rightarrow \mathbb{R}^2$ with $f_4(y)=(0,-y)$

Thank you!

Answer 1

If $D$ would be a manifold, it would be a $1-$dimensional manifold. In particular, if $U$ would be an open set of $D$ that contain $0$, then $U\setminus \{(0,0)\}$ would be an open set of $D$ with $4$ disjoints connected components, whereas, if it would have been a manifold, $U\setminus \{(0,0)\}$ would be an open set with at most $2$ disjoints connected components.