I have to show that the curve on the picture is not a submanifold. Can you tell me if my proof is correct ?
My works : I note $X$ the curve in the picture defining by $$ X = \left\{ (x,y) \mbox{ : } (x = 0,y \in \mathbb{R}) \mbox{ and } (x \in \mathbb{R},y = 0) \right\} $$
I suppose $X$ is a submanifold, so, by definiton, $ \exists U \subset \mathbb{R}^2$, a neighborhood of $(0,0) \in \mathbb{R}^2$ and $f: U \rightarrow \mathbb{R}$ a submersion in $(0,0)$ s.t $X \cap U = f^{-1}( \left\{ 0 \right\} )$.
Since $f$ is a submersion, its differential $\mathrm{d}_{(0,0)} f$ is surjective (i.e $\frac{\partial f}{\partial x}((0,0)) \neq 0$ and $\frac{\partial f}{\partial y}((0,0)) \neq 0$)
Since, $\frac{\partial f}{\partial y} \neq 0$, then by the implicit function theorem, the projection $ U \cap X \rightarrow \left\{(x,y) : y = 0\right\} $ is injective. This is a contradiction because, for all $\epsilon > 0$, points $(0,\epsilon)$ had the same image by this projection.
You are assuming that $X$ must be a $1$-dimensional submanifold. To be complete you must also show that $X$ cannot be a $2$-dimensional submanifold. Other than this, your proof works for differential manifolds.
But there are easier proofs. $X$ is not a manifold at all, even a topological one, so it cannot be a submanifold.
Note that every connected neighborhood of $(0,0)$ is broken into $4$ components when $(0,0)$ is removed. Any connected open set in $\Bbb R$ is broken into only $2$ components when a point is removed, not $4$. Any connected open set in $\Bbb R^n$ for any $n > 1$ is not disconnected by removing a single point at all.
Therefore no neighborhood of $(0,0)$ can be homeomorphic to a Euclidean space. That is, $X$ is not locally Euclidean at $(0,0)$, and thus is not a manifold.