I am struggling with the proof by contradiction that the following subset $X$ is not a submanifold of $\mathbb{R}^2$ and I would appreciate any explanation.
Let $$
X = ((-1,0] \times \{0\}) \cup (\{0\} \times [0, 1)) \subseteq \mathbb{R}^2.
$$
Suppose $X$ is a submanifold of $\mathbb{R}^2$.
Suppose $(U, \phi)$ be a local chart of $\mathbb{R}^2$ around $(0,0)$ such that $\phi|_{U \cap X}$ maps $U \cap X$ homeomorphically onto $\phi(U) \cap (\mathbb{R} \times \{0\})$. WLOG let $\phi(0,0) = 0$. Let now $(U, id)$ be a local chart of $\mathbb{R}^2$ for its $C^{\infty}$ structure. Since $D(\phi \circ id^{-1})|_0$ is singular, it follows that $\phi \circ id^{-1}$ is not a diffeomorphism. Hence the local chart $(U, \phi)$
and $(U, id)$ are not $C^{\infty}$ compatible. Contradiction. $X$ is not a $C^{\infty}$ submanifold of $\mathbb{R}^2$.
1) How come we only need to consider the dimension $1$ case? How do we know that $X$ will not be of dim $0$ and $2$ submanifold?
2) How do I prove that $D(\phi \circ id^{-1})|_0$? I can see it, because it's turning an edge to a straight line, but I was wondering how can I prove this?
Consider the parametrization $\gamma:(-\epsilon,\epsilon)\to U$ given by $\gamma(t)=(0,t)$ (We can choose $\epsilon<1$ small enough, so that $(0,t)\in U$ for $-\epsilon<t<\epsilon$). Since $\gamma'(0)=(0,1)$, we have that $$D(\phi\circ id^{-1})|_0(0,1)=\frac{d}{dt}(\phi\circ id^{-1}\circ\gamma(t))|_{t=0}=\lim_{t\to 0^+}\frac{\phi(\gamma(t))}{t}$$ But, since $\gamma(t)\in X\cap U$ for $0<t<\epsilon$, we have that $\phi(\gamma(t))\in \phi(U)\cap\mathbb{R}\times \{0\}\subset \mathbb{R}\times \{0\}.$ In particular $$D(\phi\circ id^{-1})|_0(0,1)=\lim_{t\to 0^+}\frac{\phi(\gamma(t))}{t}\in\mathbb{R}\times \{0\}$$.
A similar approach, proves that $D(\phi\circ id^{-1})|_0(1,0)\in \mathbb{R}\times \{0\}$. Since $(0,1);(1,0)$ is a basis in $\mathbb{R}^2$, this implies that the image of $D(\phi\circ id^{-1})|_0\subset\mathbb{R}\times\{0\}$ in particular is singular.