If $F:M\to N$ is a smooth embedding, then so is $dF:TM\to TN$. 
Question: I am trying to show that if $M$ and $N$ are smooth manifolds (without boundary), and
  $$F:M\to N$$
  is a smooth embedding, then the differential
  $$dF:TM\to TN,\quad dF(p,v)=(F(p),dF_p(v))$$
  is also a smooth embedding. 

In particular, this shows that an embedded submanifold of a smooth manifold gives rise to an embedded submanifold of the tangent bundle in a natural way.
It is not difficult to show that $dF$ is a smooth immersion. Indeed, it has coordinate representations of the form
$$dF(x,v)=(F(x),DF(x)v),\quad(x,v)\in \hat{U}\times\mathbb{R}^m\subseteq\mathbb{R}^m\times\mathbb{R}^m$$
so
$$D(dF)(x,v)=\begin{pmatrix}DF(x) & 0 \\ \ast & DF(x) \end{pmatrix},$$
which has full rank since $DF(x)$ has full rank. Hence, we at least have that $dF(TM)$ is a immersed submanifold of $TN$.
But now I am stuck in showing that $dF$ is a topological embedding. It is clearly injective, so the inverse
$$(dF)^{-1}:dF(TM)\to TM$$
exists. But how do you show it is continuous?
Definitions: Here "smooth" means $C^\infty$. The assumption that $F$ is a smooth embedding means that $F$ is a smooth immersion (i.e. $dF_p:T_pM\to T_{F(p)}N$ is injective at each $p\in M$) and that $F$ is a topological embedding (i.e. $F:M\to F(M)$ is a homeomorphism when $F(M)$ is given the subspace topology inherited from $TN$).
 A: Once you know that the image is an immersed submanifold, to show that $F$ is an embedding, it suffices to show that its image is an embedded submanifold; and for that it suffices to show it's embedded in a neighborhood of each point of the image. Let $p\in M$ and $q=F(p)$. Because $F$ is a smooth embedding, it is possible to choose smooth coordinate charts $(U,(x^i))$ containing $p$ and $(V,(y^i))$ containing $q$ such that $F(M)\cap V = F(U)$ and $F|_U$ has a coordinate representation of the form $F(x^1,\dots,x^m) = (x^1,\dots,x^m,0,\dots,0)$. If we let $(x^i, v_i)$ be the associated standard coordinates for $TM$ and $(y^i,w_i)$ those for $TN$, then $dF|_{TU}\colon TU\to TV$ has the coordinate representation 
$$
dF(x^1,\dots,x^m,v_1,\dots,v_m) = (x^1,\dots,x^m,0,\dots,0,v_1,\dots,v_m,0,\dots,0).
$$
The fact that $F(M)\cap V = F(U)$  guarantees that $dF(TM) \cap TV = dF(TU)$, and the coordinate representation above shows that $dF(TU)$ is an embedded $2m$-dimensional submanifold of $TV$.
