Proving that the antipodal map $F : S^2 \to S^2$ defined by $F(p) = -p$ is a diffeomorphism

Due to some reason I do not know what diffeomorphisms are (and just now learned the definition from Wikipedia), so I am very unsure about whether my current argument is correct.

Obviously $$F$$ is a bijection (with inverse as itself). Therefore I think I only need to prove that $$F$$ is differentiable on $$S^2$$. My proof is as follows:

Fix a point $$p \in S^2$$ with a corresponding chart $$x$$ (whose image contains $$p$$). I want to show that $$F \circ x$$ is differentiable, which is true trivially by $$F$$ differentiable in $$\mathbb{R}^n$$ and the regularity of $$x$$ (implying $$x$$ differentiable). Then $$F \circ x$$ differentiable. I can now conclude that $$F$$ is a diffeomorphism on $$S^2$$.

Is my argument correct?

If I understand correctly, your argument is this: because the map $$F:\Bbb R^3\to \Bbb R^3$$ given by $$F(p)=-p$$ is a smooth map and $$S^2$$ is an embedded submanifold of $$\Bbb R^3$$ such that $$F({S^2})=S^2$$, therefore $$F|_{S^2}:S^2\to S^2$$ is smooth. This is actually true in general. It is a combination (3) of two results (1) and (2) below.

(1) Let $$f:M\to N$$ be a smooth map between smooth manifolds. If $$X$$ is a submanifold (immersed or embeded) of $$M$$, then $$f|_X:X\to N$$ is also a smooth map.

To show this, we note that the embedding $$\iota:X\to M$$ is a smooth map. Since the composition of two smooth maps is smooth and $$f\circ \iota=f|_X$$, $$f|_X$$ is therefore smooth.

(2) Let $$f:M\to N$$ be a smooth map between smooth manifolds. If $$Y$$ is an embedded submanifold of $$N$$ such that $$f(M)\subseteq Y$$, then $$f:M\to Y$$ is smooth. If $$Y$$ is an immersed submanifold of $$N$$ such that $$f(M)\subseteq Y$$ and $$f:M\to Y$$ is a continuous map, then $$f:M\to Y$$ is smooth.

The proof is a bit long, so I refer you to Theorem 5.29 of John Lee's Introduction to Smooth Manifolds.

(3) Let $$f:M\to N$$ be a smooth map between smooth manifolds and $$X$$ is a submanifold (immersed or embedded) of $$M$$. If $$Y$$ is an embedded submanifold of $$N$$ such that $$f(X)\subseteq Y$$, then $$f|_X:X\to Y$$ is smooth. If $$Y$$ is an immersed submanifold of $$N$$ such that $$f(X)\subseteq Y$$ and $$f|_X:X\to Y$$ is a continuous map, then $$f|_X:X\to Y$$ is smooth.

This is simply a corollary of (1) and (2).

Here is a more hands-on proof. Let $$S^2=\big\{(X,Y,Z)\in \Bbb R^3\ :\ X^2+Y^2+Z^2=1\big\}$$. To show that $$F$$ is differentiable (smooth in this case) is to show that for an atlas $$\big\{(\phi_\alpha:U_\alpha\to V_\alpha)\ :\ {\alpha\in J}\big\}$$ of $$S^2$$, the maps $$\psi_{\alpha,\beta}:\phi_\alpha^{-1}\big(U_\alpha\cap U_\beta\cap F^{-1}(U_\alpha\cap U_\beta)\big) \to \phi^{-1}_\beta\big(U_\alpha\cap U_\beta\cap F(U_\alpha\cap U_\beta)\big)$$ are smooth, where $$\psi_{\alpha,\beta}(x)=\phi_\beta^{-1}\circ F\circ \phi_\alpha(x)$$. Recall that $$S^2$$ has an atlas $$\big\{(\phi_s:U_s\to V_s),(\phi_n:U_n\to V_n)\big\}$$ given by stereographic projections from the south pole $$s=(0,0,-1)$$ and the north pole $$n=(0,0,1)$$. To be precise, $$U_s=S^2\setminus \{s\}$$, $$U_n=S^2\setminus\{n\}$$, $$V_s=V_n=\Bbb R^2$$, $$\phi_s(X,Y,Z)=\left(\frac{X}{1+Z},\frac{Y}{1+Z}\right)$$ and $$\phi_n(X,Y,Z)=\left(\frac{X}{1-Z},\frac{Y}{1-Z}\right).$$

From the given definition of $$F$$, we can see that $$\psi_{s,n}(x,y)=(-x,-y)=\psi_{n,s}(x,y)$$ for all $$(x,y)\in \Bbb R^2$$. Likewise $$\psi_{s,s}(x,y)=\left(-\frac{x}{x^2+y^2},-\frac{y}{x^2+y^2}\right)=\psi_{n,n}(x,y)$$ for all $$(x,y,z)\in\Bbb R^2\setminus\big\{(0,0)\big\}$$. These are smooth maps.

Since $$F^{-1}=F$$, $$F^{-1}$$ is also smooth. That is, $$F$$ is a diffeomorphism on $$S^2$$.