My lecturer, to demonstrate that the projective complex space $\mathbb{P}^n(\mathbb{C})$ is Hausdorff, proves that the restriction of the quotient map to $S^{2n+1}$ is a closed map. Here is how the proof begins.
Let $$ \pi_0 := \pi_{\mid S^{2n+1}} : S^{2n+1} \to \mathbb{P}^n(\mathbb{C})$$ be the restriction of the quotient map $\pi: \mathbb{C}^{n+1}\setminus \{0\} \to \mathbb{P}^n (\mathbb{C})$. We want to prove the closedness of $\pi_0$: $\forall C$ closed set in $S^{2n+1}$ $\pi_0(C)$ is a closed set in $\mathbb{P}^n(\mathbb{C})\iff \pi_0^{-1}\left(\pi_0(C)\right)$ is closed in $S^{2n+1}$.
However, being in a quotient topology with quotient map $\pi$, proving that $\pi_0(C)$ is a closed set in $\mathbb{P}^n(\mathbb{C})$ should be equivalent to prove that $\pi^{-1}\left(\pi_0(C)\right)$ is closed in $\mathbb{C}^{n+1}\setminus \{0\}$. Therefore, my question is: does $\pi^{-1}\left(\pi_0(C)\right)$ being closed in $\mathbb{C}^{n+1}\setminus \{0\}$ implies that $\pi_0^{-1}\left(\pi_0(C)\right)$ is closed in $S^{2n+1}$ and viceversa?
This community wiki solution is intended to clear the question from the unanswered queue.
As shown in the comments, the answer is "yes". Also have a look at The restriction of a quotient map $p : X \to Y$ to a retract of $X$ is again a quotient map. This shows that $\pi_0 : S^{2n+1} \to \mathbb{P}^n(\mathbb{C})$ is a quotient map because $\pi: \mathbb{C}^{n+1}\setminus \{0\} \to \mathbb{P}^n (\mathbb{C})$ is one. Therefore the following are equivalent for $A \subset \mathbb{P}^n (\mathbb{C})$:
$A$ is closed in $\mathbb{P}^n (\mathbb{C})$.
$\pi^{-1}(A)$ is closed in $\mathbb{C}^{n+1}\setminus \{0\}$.
$\pi_0^{-1}(A)$ is closed in $S^{2n+1}$.