I'm proving a remark from this lecture note. Could you please check if my proof is fine or contains logical mistakes?
Theorem: Let $X \subseteq \mathbb R^M$ and $Y \subseteq \mathbb R^N$ be $m$- and $n$-dimensional manifolds respectively. Then $X \times Y$ is a $(m+n)$-dimensional manifold.
Proof: For $x \in X$, there exist $U_x$ open in $X$, $V_x$ open in $\mathbb R^m$, and a diffeomorphism $f_x:U_x \to V_x$ such that $x \in U_x$. For $y \in Y$, there are $U_y$ open in $Y$, $V_y$ open in $\mathbb R^n$, and a diffeomorphism $f_y : U_y \to V_y$ such that $y \in U_y$.
We define $f: U_x \times U_y \to V_x \times V_y$ by $f(a,b) = (f_x(a), f_y(b))$. Clearly, $f$ is bijective, $U_x \times U_y$ is open in $X \times Y$, and $V_x \times V_y$ is open in $\mathbb R^m \times \mathbb R^n = \mathbb R^{m+n}$.
It follows from $(a, b) \mapsto a \mapsto f_x(a)$ that the first coordinate of $f$ is smooth. Similarly, the second coordinate of $f$ is smooth. This in turn implies $f$ is smooth. Given $(c, d) \in V_x \times V_y$, we have $(c, d) \mapsto c \mapsto f_x^{-1} (c)$, so the first coordinate of $f^{-1}$ is smooth. Similarly, the second coordinate of $f^{-1}$ is smooth. Hence $f^{-1}$ is also smooth.
It follows that $f$ is indeed a diffeomorphism from $U_x \times U_y$ to $V_x \times V_y \subseteq \mathbb R^{m+n}$. This completes the proof.
I recall related definitions from the note to avoid ambiguity.
Let $U \subseteq \mathbb{R}^{N}$ be open. A map $f: U \rightarrow \mathbb{R}^{n}$ is smooth if all of its partial derivatives of all orders exist and are continuous.
Let $X \subseteq \mathbb{R}^{N}$, $Y \subseteq \mathbb{R}^{n}$, $f: X \rightarrow Y$, and $x \in X$. A map $\tilde{f}: U \rightarrow \mathbb{R}^{n}$ is a smooth extension of $f$ around $x$ if $U$ is open in $\mathbb R^N$, $x \in U$, $\tilde f$ is smooth, and $\tilde{f}$ agrees with $f$ on $U \cap X$.
Let $X \subseteq \mathbb{R}^{N}$ and $Y \subseteq \mathbb{R}^{n}$. A map $f: X \rightarrow Y$ is smooth if $f$ has a smooth extension around any $x \in X$.
A bijective smooth map $f: X \rightarrow Y$ with $X \subseteq \mathbb{R}^{N}$ and $Y \subseteq \mathbb{R}^{n}$ is a diffeomorphism if its inverse $f^{-1}: Y \rightarrow X$ is also smooth.
A subset $X \subseteq \mathbb{R}^{N}$ is a smooth $n$-dimensional manifold if $\forall x \in X$, $\exists$ a diffeomorphism $\varphi : U \to V$ such that $U$ is open in $\mathbb{R}^{n}$, $V$ is open in $X$, and $x \in V$. Then $\varphi$ is a local parameterization around $x$. The inverse $\varphi^{-1}$ is a local coordinate system (or chart) around $x$.
Let $X \subseteq \mathbb R^N$ be a smooth $n$-dimensional manifold and $x \in X$. Let $\varphi: U \to V$ be a local parameterization around $x$. The map $\mathrm d \varphi_{\varphi^{-1}(x)} : \mathbb R^n \to \mathbb R^N$ is the Fréchet derivative of $\varphi$ at $\varphi^{-1}(x)$. The tangent space of $X$ at $x$, denoted by $T_xX$, is the image of $\mathrm d \varphi_{\varphi^{-1}(x)}$, i.e., $$T_xX := \operatorname{im} \left (\mathrm d \varphi_{\varphi^{-1}(x)} \right ).$$
