Following the proof given in Milnor's Topology From A Differentiable Viewpoint:
$\require{AMScd}$
$\begin{CD}
x \in U \subseteq M @>F>> c \in V \subseteq N \\
@V\phi VV @VV\psi V \\
\phi\left(U\right) \subseteq H^{m} @>>\psi F \phi^{-1}> \psi\left(c\right) \in \psi\left(V\right)
\end{CD}$
Because $c \in N$ is a regular value of $F$, for every $x \in F^{-1}\left(c\right) \subseteq M$, there are charts $\left(U,\phi\right)$ at $x$ in $M$ and $\left(V,\psi\right)$ at $c$ in $N$ such that $\psi F \phi^{-1}: \phi\left(U\right) \subseteq H^{m} \to \psi\left(V\right) \subseteq \mathbb{R}^{n}$ is smooth, and has a regular value at $\psi\left(c\right)$.
$\begin{CD}
\phi\left(U\right) \subset W \subseteq \mathbb{R}^{m} @>G>> \psi\left(c\right) \in \mathbb{R}^{n}
\end{CD}$
Let $W$ be an open subset of $\mathbb{R}^{m}$ such that $W \cap H^{m} = \phi\left(U\right)$; and let $G:W\to\mathbb{R}^{n}$ be the smooth extension of $\psi F \phi^{-1}$ over $W$. Now, we can always choose $W$ small enough so that $G^{-1}\left(\psi\left(c\right)\right)$ does not contain any critical points (Sard's lemma; $\mathbb{R}^{m}$ is regular). Thus, $\psi\left(c\right)$ is a regular value of $G$; and by preimage theorem (for smooth manifolds), $Z := G^{-1}\left(\psi\left(c\right)\right)$ is a (m-n)-dimensional submanifold of $\mathbb{R}^{m}$.
Furthermore, since $G$ is constant over $Z$, $T_{a}Z \subseteq ker\left\{DG\left(a\right)\right\}$ for every $a \in Z$. But $DG\left(a\right):\mathbb{R}^{m}\to\mathbb{R}^{n}$ is surjective, and the rank-nullity theorem implies $dim \left(ker\left\{DG\left(a\right)\right\}\right) =$ (m-n). Thus, $T_{a}Z = ker\left\{DG\left(a\right)\right\}$.
Now, define $\pi: Z \subseteq \mathbb{R}^{m} \to \mathbb{R}$ as $\left(x_{1},\ldots,x_{m}\right) \mapsto x_{m}$.
To show that $0 \in \mathbb{R}$ is a regular value of $\pi$:
Suppose otherwise. That is, suppose $\exists$ $a \in Z \cap \partial H^{m}$ such that $d\pi_{a}:T_{a}Z \to \mathbb{R}$ is not surjective. Then, $ker \left\{d\pi_{a}\right\}$ $=$ $T_{a}Z$ $=$ $ker \left\{DG(a)\right\}$. But, we know that $ker \left\{d\pi_{a}\right\} \subseteq \mathbb{R}^{m-1} \times \left\{0\right\} = \partial H^{m}$. Thus, $ker \left\{DG(a)\right\} \subseteq \partial H^{m}$ if $0$ is not a regular value for $\pi$.
Now, since $c \in N$ is a regular value of $F|_{\partial M}$ as well, arguing as before, we can show that $\bar{G} := G|_{W\cap\partial H^{m}}$ has a regular value at $\psi\left(c\right)$. That is, for every $a \in Z\cap\partial H^{m}$, $D\bar{G}\left(a\right): \mathbb{R}^{m-1} \to \mathbb{R}^{n}$ is surjective; and by rank-nullity theorem, $dim \left(ker\left\{D\bar{G}\left(a\right)\right\}\right) = $ (m-n-1)
Finally, $ker \left\{DG(a)\right\} \subseteq \partial H^{m}$ implies $ker \left\{DG(a)\right\} = ker \left\{D\bar{G}(a)\right\}$, which is clearly false (dimension mismatch). Hence, $0$ must be a regular value for $\pi$.
Since $0 \in \mathbb{R}$ is a regular value for $\pi$, $\left\{z \in Z | \pi\left(z\right) \geq 0\right\}$ $=$ $\phi\left(U \cap F^{-1}\left(c\right)\right)$ is a manifold with boundary $\left\{z \in Z | \pi\left(z\right) = 0\right\}$ $=$ $\phi\left(U \cap F^{-1}\left(c\right) \cap \partial M \right)$.
$\phi$ being a diffeomorphism, $U \cap F^{-1}\left(c\right)$ is a manifold with boundary $U \cap F^{-1}\left(c\right) \cap \partial M$. Observing that this is true for every $x \in F^{-}\left(c\right)$ completes the proof.
If you allow your manifolds to have boundary, then no: consider $M=[0,1)$, $N=\mathbb{R}^2$, and let $F$ be an immersion whose image is a topological circle which has a non-smooth corner at $F(0)$ (with $F(t)$ approaching $F(0)$ as $t\to 1$).
If you do not allow boundaries, then yes, by invariance of domain. Invariance of domain implies that any continuous bijection between topological manifolds without boundary is an open map and thus a homeomorphism.* In particular, $F:M\to F(M)$ must be a homeomorphism so $F:M\to N$ is a smooth embedding.
*Note that if $f:X\to Y$ is a continuous bijection between (second-countable) nonempty topological manifolds then $X$ and $Y$ must have the same dimension, and so invariance of domain applies. This follows from the fact that $\mathbb{R}^n$ does not embed in $\mathbb{R}^m$ if $n>m$, which is another consequence of invariance of domain (restrict such an embedding to an $m$-dimensional subspace of $\mathbb{R}^n$ and apply invariance of domain to get a contradiction). This immediately implies that $\dim X\leq \dim Y$. On the other hand, $f$ is an embedding when restricted to each compact subset of $X$, and the images of these compact subsets must have empty interior if $\dim X<\dim Y$. Since $X$ is $\sigma$-compact this implies $f$ cannot be surjective by the Baire category theorem.
Best Answer
Assume $M$ connected. Note that $A=Im(f)$ is therefore connected. Let $a\in A$, $f'(a)$ is a linear projection on $T_aM$. As the rank of a projection is a contious function on the set of projection (it is the trace), along $A$ the rank of $f'$ is constant $r$. Now, let us check that it is constant in a neigbourhood of $A$ . As $f\circ f= f$, $f'(f(p))\circ f'(p)= f'(p)$, and $f'(f(p))$ induces a isomorphism on $Im ( f'(p))$, so that the rank of $f'(p)$ is less than or equal to $r$. But the rank can only increase in the neigbourhood of $A$, so it is constant. If you know the "constant rank theorem" you are done. If not let us argue as follows.
Now choose a point $a\in A$. In order to study the image of $f$ near this point, let us choose coordinates $x_1,...,x_n$ near $a$ such that $f'(a)$ is the projection onto $x_{r+1}==x_n=0$, and let us consider the new system of coordiantes $y_i=x_i\circ f$ for $i\leq r$, $y_i=x_i$ if $i> r$. There are functions $F_j, j>r$ such that $f(y_1,..y_n)=(y_1,,..,y_r, F_{r+1},...F_{n})$ Writing that the rank of $f'$ is $r$ we see that ${\partial F_j \over \partial y_k }=0$ if $k>r$. Or $f(y_1,...y_n)= (y_1,..,y_r, F_1(y_1,..) ..)$. So near $a$ the image of $F$ is just the graph of the function $F=(F_{r+1},...F_n$), a submanifold.