Every embedding gives rise to an embedded submanifold? I call regular (or embedded) submanifold of a manifold $M$ a subset  $i:X \subset M$ endowed with a submanifold structure such that the inclusion map $i$ is an embedding.
I call embedding an immersion that is also an homeomorphism onto its image.
I call immersion a smooth map of constant rank equal to the dimension of its domain.
It can be proven that a smooth constant rank map is injective only if its rank is equal to the dimension of its domain. Hence any smooth constant rank injective map must be an immmersion.
An inclusion is homeomorphic onto its image iff on $X$ I have the subspace topology.
Moreover, since the inclusion is injective, to prove it is an immersion it suffices to show that it is smooth and of constant rank.
Hence, I can rephrase the definition of submanifold as:

A submanifold $X \subset M$ is a subset of $M$ endowed with the subspace topology and with a manifold structure such that the inclusion $i:X \subset M$ is smooth and of constant rank.

question 1

Is this rephrasing correct?

question 2

If so, what is an example of an inclusion which is not smooth? What is an example of an inclusion which is not of constant rank?

question 3
I read that for any embedding $f:N \to M,$ we have that $f(N)\subset M$ is a regular submanifold of $M.$
I understand that if we factorize $f$ as $$N\xrightarrow{\tilde{f}}f(N)\xrightarrow{i}M$$ I have that $f$ smooth of constant rank $\implies$ $\tilde{f}$ smooth of constant rank, because I am just changing the codomain.

But how to show that this must imply $i$ smooth of constant rank too?

I believe we should have a statement of the form

$h \circ g$ smooth of constant rank and $g$ smooth of constant rank
$\implies$ $h$ smooth of constant rank.
Is this correct?

 A: First of all, my favorite definition of a smooth submanifold is "none of the above," but the one I give here. Even with your definition, working with $f$ and $i$ separately only complicates things. You can say that a subset $X$ of a smooth manifold $M$ is smooth submanifold if there exists a smooth manifold $N$ and an immersion $f: N\to M$ such that $f(N)=X$ and $f: N\to M$ is a topological embedding, i.e. is a homeomorphism $N\to X$ (where $X$ is given the subspace topology which is assumed by default in topology).
Question 1: Yes, this is correct.
Question 2. (a) An inclusion which is not smooth: Consider the graph $\Gamma$ of any continuous function of one variable which is not everywhere smooth, e.g.  $y=|x|$. Then $\Gamma$ cannot be given structure of a smooth submanifold. However, the map $\phi: x\mapsto (x, |x|)$ is a homeomorphism from ${\mathbb R}$ to $\Gamma$. Hence, $\phi$ can be used to define a smooth manifold structure on $\Gamma$ (use $\phi^{-1}$ as the single chart).
(b) An inclusion which is not of constant rank. The simplest example I know comes from the map $\phi: {\mathbb R}\to {\mathbb R}$, $\phi(x)=x^3$. Then $X=M= {\mathbb R}$ (as sets!), but, as smooth manifolds I will use the structure on $X$ such that $\phi$ is a diffeomorphism. Then the inclusion map $i: X\to M$ does not have constant rank.
If you like more interesting examples, consider $X\subset {\mathbb R}^2$ equal to the solution set of the equation $y^3=x^2$. Take $f: {\mathbb R}\to X$ given by
$$
f(t)=(t^3, t^2). 
$$
Then $f$ is smooth, is a homeomorphism to its image but does not have constant rank. I leave it to you to convert this into an example when the inclusion map $i$ is smooth, is topological embedding, but does not have constant rank.
Question 3. Yes, you read it correctly. A hint for the proof: Suppose that
you have smooth maps of manifolds
$$
A \stackrel{f}{\to} B \stackrel{g}{\to} C,
$$
where $f$ is a diffeomorphism. Suppose that the map $h:= g\circ f$ is an immersion. Then $g$ is also an immersion.
