[proofplan]
The proof is the formal unpacking of the definition of [projective space](/page/Projective%20Space) as nonzero vectors modulo nonzero scalar multiplication. First we check that the nonzero affine cone is stable under the scalar action, because multiplying a representative vector by a nonzero scalar does not change its projective point. Then we define the map from scalar orbits to projective points and verify that it is independent of the chosen representative. Finally, surjectivity follows by choosing a homogeneous coordinate representative of a point of $X$, and injectivity follows from the defining [equivalence relation](/page/Equivalence%20Relation) on projective coordinates.
[/proofplan]
[step:Restrict the scalar action to the nonzero affine cone]
Define
\begin{align*}
V := C(X)\setminus\{0\}.
\end{align*}
Let
\begin{align*}
\alpha: k^\times \times V \to k^{n+1}\setminus\{0\},\quad (\lambda,a)\mapsto \lambda a
\end{align*}
be the restriction of ordinary scalar multiplication on $k^{n+1}$. We show that $\alpha$ has image in $V$.
Take $\lambda \in k^\times$ and $a=(a_0,\dots,a_n)\in V$. Since $a\in C(X)\setminus\{0\}$, the projective point $[a_0:\dots:a_n]$ lies in $X$. Also $\lambda a\ne 0$, because $\lambda\ne 0$ and $a\ne 0$. In projective space,
\begin{align*}
[\lambda a_0:\dots:\lambda a_n]=[a_0:\dots:a_n].
\end{align*}
Hence $[\lambda a_0:\dots:\lambda a_n]\in X$, so $\lambda a\in C(X)\setminus\{0\}=V$.
The usual scalar multiplication identities satisfy
\begin{align*}
1\cdot a=a
\end{align*}
for every $a\in V$, and
\begin{align*}
\lambda\cdot(\mu\cdot a)=(\lambda\mu)\cdot a
\end{align*}
for every $\lambda,\mu\in k^\times$ and $a\in V$. Therefore scalar multiplication restricts to an action of $k^\times$ on $V$.
[/step]
[step:Define the orbit map and prove it is well defined]
Let $(C(X)\setminus\{0\})/k^\times=V/k^\times$ denote the set of $k^\times$-orbits in $V$. For $a=(a_0,\dots,a_n)\in V$, denote its orbit by
\begin{align*}
k^\times a:=\{\lambda a:\lambda\in k^\times\}.
\end{align*}
Define
\begin{align*}
\Phi: V/k^\times \to X,\quad k^\times a\mapsto [a_0:\dots:a_n].
\end{align*}
This formula lands in $X$ because $a\in V\subset C(X)$ and $a\ne 0$, so the definition of $C(X)$ gives $[a_0:\dots:a_n]\in X$.
To prove well-definedness, suppose $a,b\in V$ have the same orbit. Then there exists $\lambda\in k^\times$ such that $b=\lambda a$. Writing $a=(a_0,\dots,a_n)$ and $b=(b_0,\dots,b_n)$, this means $b_i=\lambda a_i$ for every $i\in\{0,\dots,n\}$. Therefore
\begin{align*}
[b_0:\dots:b_n]=[\lambda a_0:\dots:\lambda a_n]=[a_0:\dots:a_n].
\end{align*}
Thus the value of $\Phi(k^\times a)$ depends only on the orbit $k^\times a$, not on the chosen representative $a$.
[guided]
We now define the candidate quotient map and check the point where quotient constructions can fail: independence of representatives. Let
\begin{align*}
V := C(X)\setminus\{0\}.
\end{align*}
The quotient $V/k^\times$ is the set whose elements are scalar orbits. For a vector $a=(a_0,\dots,a_n)\in V$, its orbit is
\begin{align*}
k^\times a:=\{\lambda a:\lambda\in k^\times\}.
\end{align*}
The intended map is
\begin{align*}
\Phi: V/k^\times \to X,\quad k^\times a\mapsto [a_0:\dots:a_n].
\end{align*}
First, the displayed formula has its values in $X$. Indeed, $a\in V$ means $a\in C(X)$ and $a\ne 0$. By the definition of the affine cone $C(X)$, every nonzero vector in $C(X)$ represents a projective point of $X$. Hence $[a_0:\dots:a_n]\in X$.
The essential issue is well-definedness. An element of $V/k^\times$ is not a single vector but an orbit of vectors, so two different representatives of the same orbit must produce the same projective point. Suppose $a,b\in V$ represent the same orbit. By the definition of orbit equality for a [group action](/page/Group%20Action), there exists $\lambda\in k^\times$ such that $b=\lambda a$. If $a=(a_0,\dots,a_n)$ and $b=(b_0,\dots,b_n)$, then $b_i=\lambda a_i$ for each coordinate index $i\in\{0,\dots,n\}$. The defining equivalence relation for projective coordinates identifies nonzero scalar multiples, so
\begin{align*}
[b_0:\dots:b_n]=[\lambda a_0:\dots:\lambda a_n]=[a_0:\dots:a_n].
\end{align*}
Thus changing representatives inside the same $k^\times$-orbit does not change the output of $\Phi$. This proves that $\Phi$ is a well-defined map from the quotient set $V/k^\times$ to $X$.
[/guided]
[/step]
[step:Choose homogeneous representatives to prove surjectivity]
Let $x\in X$. Since $x$ is a point of $\mathbb{P}^n(k)$, there exists a nonzero vector $a=(a_0,\dots,a_n)\in k^{n+1}\setminus\{0\}$ such that
\begin{align*}
x=[a_0:\dots:a_n].
\end{align*}
Because $x\in X$, the defining condition for the affine cone gives $a\in C(X)$. Since $a\ne 0$, we have $a\in V$. Therefore $k^\times a\in V/k^\times$, and
\begin{align*}
\Phi(k^\times a)=[a_0:\dots:a_n]=x.
\end{align*}
Hence $\Phi$ is surjective.
[/step]
[step:Use equality of projective points to prove injectivity]
Suppose $a=(a_0,\dots,a_n)\in V$ and $b=(b_0,\dots,b_n)\in V$ satisfy
\begin{align*}
\Phi(k^\times a)=\Phi(k^\times b).
\end{align*}
By the definition of $\Phi$, this means
\begin{align*}
[a_0:\dots:a_n]=[b_0:\dots:b_n].
\end{align*}
Equality in projective space means precisely that there exists $\lambda\in k^\times$ such that
\begin{align*}
b=\lambda a.
\end{align*}
Thus $b\in k^\times a$, so $k^\times b=k^\times a$ as orbits in $V$. Therefore $\Phi$ is injective.
[/step]
[step:Identify the projective variety with the scalar quotient as a set]
The map $\Phi:V/k^\times\to X$ is both surjective and injective, and it is well defined. Hence $\Phi$ is a bijection. Since $V=C(X)\setminus\{0\}$, this gives the set-theoretic identification
\begin{align*}
X \cong (C(X)\setminus\{0\})/k^\times.
\end{align*}
This is the asserted quotient description of $X$ as the set of nonzero vectors in its affine cone modulo nonzero scalar multiplication.
[/step]
