Below is my proof that $P(\mathbb{C}^{n+1})$ complex projective space of dimension $n$ is hausdorff with the usual topology. I would appreciate it if someone could verify whether it is sound or not.
Preamble:
Define $\mathbb{P}^n = W / \tilde{}$, where $W$ is $\mathbb{C}^{n+1}$ with $0$ removed and $\tilde{}$ is the usual equivalence relation given by $x \tilde{} y$ if there exists some non zero $\lambda \in \mathbb{C}^{*}$ such that $x = \lambda y$. If $x \in W$, denote the equivalence class of $x = (x_0,x_1,...,x_n)$ by $[x_0,...,x_n]$.
The topology on $\mathbb{P}^n$ is the usual quotient topology.
Define $U_i = \{[x_0,...,x_n] \in \mathbb{P}^n | x_i \neq 0 \}$ and consider the bijection $\phi_i : \mathbb{C}^n \rightarrow U_i$, $\phi_{i}((y_1,...,y_n)) = [y_1,...,y_i,1,y_{i+1},...,y_n]$ with inverse $\psi_{i}([x_0,...,x_n]) = (\frac{x_0}{x_i},...,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},...,\frac{x_n}{x_i})$. It is a fact that the quotient topology (the final or largest topology on $\mathbb{P}^n$ so that the quotient map is continuous) is equivalent to the final topology on $\mathbb{P}^n$ so that each $\phi_i$ is continuous when $U_i$ has the subspace topology, in other words $U \subset \mathbb{P}^n$ is open iff $\phi_{i}^{-1}(U \cap U_i) \subset \mathbb{C}^n$ is open for each $i$. In particular, this gives that each $U_i$ is open in $\mathbb{P}^n$ as one can check that $\phi_{i}^{-1}(U_i \cap U_j)$ is homeomorphic to $\{(x_1,x_2,...,x_n) | x_1 \neq 0, (x_2,...,x_n) \in \mathbb{C}^{n-1} \}$ which is certainly open in $\mathbb{C}^{n}$.
It is also clear that $\mathbb{P}^n = \bigcup_{i=0}^n U_i$, where each $U_i$ is homeomorphic to $\mathbb{C}^n$ by the above described bijection ($\phi_{i}$)
I want to prove that $\mathbb{P}^n$ with the above described topology (equivalent to the usual quotient topology on $\mathbb{P}^n$) is hausdorff.
Proof:
First, if $x \neq y \in \mathbb{P}^n$ are such that $x,y \in U_j$, for some $j = 0,...,n$, then since $U_j$ is homeomorphic to $\mathbb{C}^n$, it is hausdorff, and thus there are open sets in $U_j$, of the form $U_j \cap U_x$ and $U_j \cap U_y$ that are disjoint and separate $x$ and $y$.
In the event $x \neq y$ do not both belong to any of the $U_i$, WLOG we may assume that $x = [1,x_1,...,x_k,0,...,0]$ and $y = [0,0,...,0,1,y_2,...,y_{n-k}]$, where each $x_i \neq 0$ and $y_j \in \mathbb{C}$. (Of course if $k \leq 0$ then $x = [1,0,...,0]$).
Now let $B_{\epsilon}(M) = A_{\epsilon}$ and $B_{\epsilon}(P) = B_{\epsilon}$ be two open balls of radius $\epsilon$ centered at $M = (1,x_1,...,x_k,0,...,0)$ and $P = (0,0,...,0,1,y_2,...,y_{n-k})$ respectively in $\mathbb{C}^{n+1}$.
Claim that for $\epsilon$ sufficiently small, $q(A_{\epsilon}) \cap q(B_{\epsilon}) = \emptyset$, where $q: \mathbb{C}^{n+1} \rightarrow \mathbb{P}^{n}$, $q((x_0,...,x_n)) = [x_0,...,x_n]$ is the usual quotient map.
Suppose not. Then for all $\epsilon > 0$, there exists some $X_{\epsilon} \in A_{\epsilon}$ and non zero $\lambda_{\epsilon} \in \mathbb{C}^{*}$ such that $\lambda_{\epsilon}X_{\epsilon} \in B_{\epsilon}$. This immediately gives $lim_{\epsilon \rightarrow 0} \lambda_{\epsilon}X_{\epsilon} = P$. It is clear then, as $X_{\epsilon}$ is itself bounded, that $\lambda_{\epsilon}$ in fact is as well. We thus see that $\lambda_{\epsilon}X_{\epsilon} - \lambda_{\epsilon}M \rightarrow 0$ as $\epsilon \rightarrow 0$, and in particular this implies that $\lambda_{\epsilon}M \rightarrow P$ as $\epsilon \rightarrow 0$ which is impossible.
Thus for some $\epsilon > 0$ we must have $q(A_{\epsilon}) \cap q(B_{\epsilon}) = \emptyset$, and since we are working in the quotient topology on $\mathbb{P}^n$, and $x \in q(A_{\epsilon})$ and $y \in q(B_{\epsilon})$ we have found open sets that are disjoint separating our points.
This concludes the proof.