Let $S^n := \{x \in \mathbb R^{n+1} \mid x_1^2+ \cdots+x_{n+1}^2 = 1\}$. Then $S^n$ is a smooth $n$-dimensional manifold This is my first proof in differential geometry. Could you have a check if I  understand the concepts correctly.
Theorem: Let $S^n := \{x \in \mathbb R^{n+1} \mid x_1^2+ \cdots+x_{n+1}^2 = 1\}$. Then $S^n$ is a smooth $n$-dimensional manifold.
Proof: Let $S^n_{+i} := \{x \in S^n \mid x_{i} > 0\}$ and $S^n_{-i} := \{x \in S^n \mid x_{i} < 0\}$ for $i = 1,\ldots, n+1$, and ${\mathbb B}_n (0, 1)$ be the open ball with center $0 \in \mathbb R^n$ and radii $1$. Notice that

*

*$S^n_{+i}$ and $S^n_{-i}$ are open in $S^n$ w.r.t. the relative topology of $S^n$. This is due to the strict inequality signs "$<$" and "$>$".


*Each $x \in S^n$ belongs to $S^n_{+i}$ or $S^n_{+i}$ for some $i$.
Consider $$\varphi_{+i} : S^n_{+i} \to {\mathbb B}_n (0, 1) \quad \text{and} \quad \varphi_{-i} : S^n_{-i} \to {\mathbb B}_n (0, 1),$$ where $\varphi_{+i} (x)$ and $\varphi_{-i} (x)$ are defined be removing the $i$-th coordinate from $x$. Then $\varphi_{+i}$ and $\varphi_{-i}$ are diffeomorphisms with inverses $\varphi_{+i}^{-1}$ and $\varphi_{-i}^{-1}$ respectively. Here $\varphi_{+i}^{-1} (y)$ and $\varphi_{-i}^{-1} (y)$ are defined by inserting the components $1 - \sqrt{y_1^2+ \cdots + y_{n}^2}$ and $-1 + \sqrt{y_1^2+ \cdots + y_{n}^2}$ respectively between $y_{i-1}$ and $y_{i}$.
Hence we can associate $x \in S^n$ with $\varphi_{+i}$ if $x \in S^n_{+i}$ or with $\varphi_{-i}$ if $x \in S^n_{-i}$. This completes the proof.
 A: This could not fit as a comment, so it became an answer.
"Psychologically" it is good to think of $M:=S^n$ defined as a set by
$$
M=\Big\{\ x\in \Bbb R^{n+1}\ :\ \|x\|_2=1\ \Big\}
$$
and organized as topological space by using the cited subspace topology as being something "amorph". (Maybe something visualized as a tar roof.) As it comes so far, we have no smooth structure. So far we have nothing. We have to organize $M$ as a smooth manifold. "Smooth" is usually taken as synonym for "of class $\mathcal C^\infty$". So the problem, the claim also has to give an atlas, when saying "show it is a smooth manifold". Or else just say it can be organized (after given the topological structure) as such. (But the topological structure is also only implicitly given in the claim...)
Now regarding the proof.
The proof introduces maps $\varphi_{+i}$ and $\varphi_{-i}$ from open "pieces" $U_{+i}:=S^n_{+i}$ and
$U_{-i}:=S^n_{-i}$
inside the (so far) "amorph" structure to different copies of the unit ball in the $n$-dimensional space.
The wording "are diffeomophisms" for these maps is at this point, during the proof, premature. We are allowed to work only in the category of topological spaces at this point, maps are not knowing any differentiable structure just now. They are but homeomorphisms, invertible maps in the category of topological spaces, and this is all we need.
Now we want to insure also a "(smooth) differentiable structure". This is done by using the "model" of a "(smooth) differentiable structure" taken from open pieces in $\Bbb R^n$. So we are allowed to use the word diffeomorphism only for pieces inside the "known" $\Bbb R^n$. We use the charts $\varphi_{\pm i}$ to obtain maps between "different $n$-balls" $\Bbb B(0,1)=\Bbb B_n(0,1)$ (of radius one centered in zero) inside $\Bbb R^n$. These are the red maps in the diagram below. (Here i cannot draw oblique arrows...)
$\require{AMScd}$
$$
\begin{CD}
U_{i+}\cap U_{j+} @= U_{i+}\cap U_{j+}\\
@V {\operatorname{inclusion}} V V @VV {\operatorname{inclusion}} V\\
U_{i+} @. U_{j+}\\
@V \varphi_{i+} V V @VV \varphi_{j+} V\\
\Bbb R^n\supset\operatorname{Image}(\varphi_{i+})\qquad  @>> {\color{red}{\varphi_{j+}|\ \circ\ \varphi_{i+}|^{-1}}} > \qquad\operatorname{Image}(\varphi_{j+}) \subset \Bbb R^n \\
@V \operatorname{inclusion} V V @VV  \operatorname{inclusion} V\\
\Bbb R^n\supset\Bbb B(0,1)\qquad @. \qquad\Bbb B(0,1)\subset \Bbb R^n
\end{CD}
$$
There are of course further such compositions of restrictions (that vertical bar) of these maps / charts, take instead of the two plus signs all $\pm$-possibilities.
It remains now to show that the red maps are smooth. The proof starts first at this point... One has maybe to compute compositions, observe that componentwise only "radicals and polynomials" are involved. The radicals are considered on expression that do no vanish.

After all, we have a smooth manifold, and we can speak of smooth maps between smooth manifolds. Also in this case, the "smoothness" test is done in "open affine pieces" (i.e. open sets inside real euclidean spaces of the dimensions related to the dimension of the involved manifolds).
