Let $M \subseteq \mathbb{R}^n$ be a smooth $k$-dimensional submanifold and let $A \subseteq \mathbb{R}^n$ be an open subset with $M \subseteq A$. Show that if $\psi$ is a diffeomorphism from $A \rightarrow B \subseteq \mathbb{R}^n$, then the image $\psi(M)$ is a (smooth) submanifold.
My approach: Since $M$ is a submanifold, for $x \in M$, there exist open subsets $U,V$ with $x \in U$ and a diffeomorphism $\phi: U \rightarrow V$, such that (*)$\phi(M \cap U)=V \cap \mathbb{R}^k$.
Let $x \in \psi(M)$, then $\psi^{-1}({x}) \in M$. Since $M$ is a submanifold,for $\psi^{-1}(x)$ there exists a $U,V$ and $\phi$ s.t. as in the section above. Now $x \in \psi(U)$, where $\psi(U)$ is open. If we apply $\phi^{-1}$ to (*) we get $M \cap U$= $\phi^{-1}(V \cap \mathbb{R}^k)$. Applying $\psi$ yields $\psi(U \cap M)=\psi (\phi^{-1}(V \cap \mathbb{R}^k))$, since $\psi, \phi$ are bijective we get $\psi(U) \cap \psi(M)=\psi(\phi^{-1}(V)) \cap \psi(\phi^{-1}(\mathbb{R}^k))$. And Thus $\psi(A)$ is a submanifold.
Edit:
My Questions are: Is this correct? Are there other ways to show this?
The proof is by-and-large fine, but one thing is fishy: You do not make use of the assumption that $A$ is open.
This I believe comes about as follows: You simply say that "there exist open subsets $U$, $V$ such that..." but you don't actually tells us what $U$ and $V$ are open subsets of (if I were to grade this, I would definitely deduct a point for that). The definition/result you want to use there I believe reads something like "there exist open subsets $U, V \subseteq \mathbb{R}^n$ such that...," and now you have a slight problem: $\psi$ is only defined on $A$, but $U$ need a priori not be contained in $A$! But of course we can replace $U$ with $A \cap U$ (which is again an open subset of $\mathbb{R}^n$ since $A$ is open) and $V$ with $\phi(A \cap U)$, and then the rest of your argument goes through.
As for other ways to show this, since this is a pretty elementary fact most proofs are going to be about the same, but I do want to present another proof as follows: We take as definition that a submanifold $M \subseteq \mathbb{R}^n$ of dimension $k$ is a subspace with the property that for each $x \in M$ there exists an open neighborhood $U \subseteq M$ of $x$ and a diffeomorphism $\phi\colon V \to U$, called a parametrization of $M$ around $x$ from an open subset $V \subseteq \mathbb{R}^k$ (if this differs from the definition you're used to, proving that the two definitions are equivalent may be a good exercise :).
Thus, given $x \in M$ and $V, U, \phi$ as just defined, first note that $\psi(U)$ is an open neighborhood around $\psi(x)$ since if $U = U' \cap M$ for some $U'$ open in $\mathbb{R}^n$ (this is the definition of open set in the subspace topology) then $\psi(U) = \psi(M \cap U' \cap A) = \psi(M) \cap \psi(U' \cap A)$ which is open in $\psi(M)$ (using openness of $A$), so we conclude that $\psi \circ \phi\colon V \to \psi(U)$ is a parametrization around $\psi(x)$. Since $x$ was arbitrary and $\psi$ is surjective we conclude that every point of $\psi(M)$ has such a parametrization and therefore that $\psi(M)$ is a submanifold, q.e.d.
The reason I wanted to to write this out is to make the point that even though this proof is in substance not too different from yours, it trades off symbols and symbolic manipulation for words, which (to me, at least) makes for a more readable result.