Is any immersed manifold also embedded? Let $M$ be a smooth manifold. To establish my criteria let me define immersed and embedded submanifolds:


*

*A subset $N\subset M$ is an immersed submanifold when $N$ is itself a manifold and $\iota:N\rightarrow M$ is an (injective) immersion.

*In the same way, a subset $N\subset M$ is an embedded submanifold if the inclusion is an embedding. 
By theorem 5.8 of John M. Lee's book Introduction to smooth manifolds, any subset $N$ of a smooth manifold $M$ such that each point $p\in N$ is contained in the domain of a chart $(U,\varphi)$ of $M$ verifying
$$
\varphi(U\cap N)= V \times \{ c \},
$$
for  $V\subset\mathbb R^k$ open and $c\in \mathbb R^{n-k}$ constant, is a topological submanifold of $M$ and it admits a smooth structure making it into a $k$-dimensional embedded submanifold. Conversely, any embedded submanifold exhibits such a property.
On the other hand, Proposition 5.22 of the same book states that if $N\subset M$ is an immersed submanifold then, for each point $p\in N$ there is an open neighbourhood (with respect to $N$) $U'$ of $p$ such that $U$ is an embedded submanifold. However, if $U'$ is an embedded submanifold, by the previous result, there exists a chart $(U,\varphi)$ of $M$ such that $p\in U$ and 
$$
\varphi(U'\cap U)=V \times \{ c\} ,
$$
for $V$ and $c$ as before. On the other hand, $U'=N\cap U''$, for some open neighbourhood (with respect to $M$) $U''$ of $p$. Then, if we set $\tilde U=U\cap U''$ and $\tilde\varphi = \varphi|_{\tilde U}$, we have found a chart of $M$ with $p\in \tilde U$ and such that
$$
\tilde\varphi(N\cap \tilde U)= V\times \{c \} .
$$
Then, again using theorem 5.8, we deduced that $N$ is an embedded submanifold. Maybe, warned by Lee's remark after proposition 5.22, the embedded structure of $N$ may or may not agree with the immersed structure. What I am saying is that for a subset $N$ being an immersed submanifold is such a nice property that the same set can be endowed with an embedded structure too.

Question. Am I right? Does any immersed submanifold admit a (possibly different) structure of embedded submanifold? If not, what is wrong with my previous reasoning?

Remark. Notice that being an immersed submanifold is not a trivial condition, i.e. not every subset can be endowed with such a strcture. For instance, the boundary of a square in $\mathbb R^2$ (Problem 5-9 of Lee's book). Then, not every set can be realised as an embedded submanifold.
 A: 
On the other hand, $U'=N\cap U''$, for some open neighbourhood (with respect to $M$) $U''$ of $p$.

This is false.  In the definition of an immersed submanifold, there is no assumption that $N$ has the subspace topology from $M$; the only assumption is that the set $N$ is given some manifold structure such that the inclusion map becomes an immersion.  Indeed, by definition, if $N$ did have the subspace topology, then $\iota$ would be an embedding, not just an immersion, since the only difference between and embedding and an injective immersion is whether the topology on the domain is the same as the subspace topology on the image.
For a very simple explicit counterexample, let $M=\mathbb{R}$ and let $N=\mathbb{Q}$ with the discrete topology (and its unique structure of a smooth $0$-dimensional manifold).  Then the inclusion $\iota:N\to M$ is an immersion, but if you give the subspace topology then it certainly is not a manifold.
A: Here's a counterexample I just posted somewhere else. Take $F: \mathbb{R}\to \mathbb{R^2}$ defined by $$F(t)=\left(2\cos\left(t-\frac{\pi}{2}\right), 2\sin\left(t-\frac{\pi}{2}\right)\right)$$ Then $F(\mathbb{R})$ is an immersed submanifold of $\mathbb{R^2}$ but not an embedded submanifold of $\mathbb{R^2}$.
It looks like this and goes through the origin twice as the lemniscate loops around itself:

