I have seen a couple questions involving the composition of smooth maps is smooth, but the proof I came up with does not look quite like the proofs given and I want to know if there is anything wrong with my proof.
Let $X \subset R^k$, $Y \subset R^l$, and $Z \subset R^m$. Claim: If $f : X \to Y$ and $g : Y \to Z$ are smooth, then their composition $g \circ f$ is smooth.
We have that $f$ is smooth. Therefore, for any $x \in X$, $\exists U \subset R^k$ that contains $x$ and there exists $F : U \to R^l$ that coincides with $f$ in $U \cap X$. Notice that $f(x) \in Y$. As $f(x) \in Y$ and $g$ is smooth we know that for any $x \in X$, $\exists W \subset R^l$ containing $f(x)$ and $\exists G : W \to R^m$ such that $G$ coincides with $g$ in $W \cap Y$. As a result, we can guarantee $G \circ F : U \to R^m$, therefore, $g \circ f$ must be smooth as there exists an appropriate open set $U$ and corresponding function $G \circ F$.