From Lee's Introduction to Smooth Manifolds:
Proposition 5.2 (Images of Embeddings as Submanifolds). Suppose $M$ is a smooth manifold with or without boundary, $N$ is a smooth manifold, and $F: N \to M$ is a smooth embedding. Let $S = F(N)$. With the subspace topology, $S$ is a topological manifold, and it has a unique smooth structure making it into an embedded submanifold of $M$ with the property that $F$ is a diffeomorphism onto its image.
Proof. If we give $S$ the subspace topology that it inherits from $M$, then the assumption that $F$ is an embedding means that $F$ can be considered as a homeomorphism from $N$ onto $S$, and thus $S$ is a topological manifold. We give S a smooth structure by taking the smooth charts to be those of the form $(F(U), \phi \circ F^{-1})$, where $(U, \phi)$ is any smooth chart for $N$; smooth compatibility of these charts follows immediately from the smooth compatibility of the corresponding charts for $N$. With this smooth structure on $S$, the map $F$ is a diffeomorphism onto its image (essentially by definition)...
Why is $F^{-1}$ smooth?
This holds because smooth embeddings are diffeomorphisms onto their images, so the inverse (suitably restricted to be well-defined) is smooth by definition. See the answer here https://math.stackexchange.com/a/321364/458920