Let $G$ be a compact Hausdorff group.
Let $X$ be a locally compact Hausdorff space.
Suppose $G$ acts continuously on $X$.
We will prove that the orbit space $X/G$ is Hausdorff.
The proof is due to Bourbaki.
As an immediate corollary, projective spaces over $\mathbb{R}$ and $\mathbb{C}$ are Hausdorff.
Definition 1
Let $X, Y$ be locally compact Haudsorff spaces.
Let $f\colon X \rightarrow Y$ be a continuous map.
If $f^{-1}(K)$ is compact for every compact subset $K$ of $Y$, $f$ is called a proper map.
Lemma 1
Let $X$ be a topological space.
Let $(A_i)_{i\in I}$ be a family of subsets of $X$.
Suppose $X = \bigcup_i int(A_i)$, where $int(A_i)$ is the interior of $A_i$.
Let $B$ be a subset of $X$.
Suppose $B \cap A_i$ is closed in $A_i$ for every $i \in I$.
Then $B$ is closed in $X$.
Proof:
Let $C = X - B$.
Since $C \cap A_i = A_i - (B \cap A_i)$, $C \cap A_i$ is open in $A_i$.
Hence there exists an open subset $V_i$ of $X$ such that $C \cap A_i = V_i \cap A_i$.
Then $C \cap int(A_i) = (C \cap A_i) \cap int(A_i) = (V_i \cap A_i) \cap int(A_i)
= V_i \cap int(A_i)$.
Hence $C \cap int(A_i)$ is open in $X$.
Since $X = \bigcup int(A_i)$, $C = \bigcup (C \cap int(A_i))$.
Hence $C$ is open in $X$.
Hence $B$ is closed in $X$.
QED
Lemma 2
Let $X, Y$ be topological spaces.
Let $f\colon X \rightarrow Y$ be a continuous map.
Let $(A_i)_{i\in I}$ be a family of subsets of $Y$.
Suppose $Y = \bigcup_i int(A_i)$.
Let $f_i\colon f^{-1}(A_i) \rightarrow A_i$ be the restriction of $f$ for each $i \in I$.
Suppose $f_i$ is a closed map for every $i \in I$.
Then $f$ is a closed map.
Proof:
Let $B$ be a closed subset of $X$.
Since $f(B) \cap A_i = f_i(B \cap f^{-1}(A_i))$, $f(B) \cap A_i$ is closed in $A_i$ for every $i \in I$.
Hence $f$ is a closed map by Lemma 1.
QED
Lemma 3
Let $X, Y$ be locally compact Hausdorff spaces.
Let $f\colon X \rightarrow Y$ be a proper map.
Then $f$ is a closed map.
Proof:
There exists a family $(A_i)_{i\in I}$ of compact subsets of $Y$ such that $Y = \bigcup_i int(A_i)$.Let $f_i\colon f^{-1}(A_i) \rightarrow A_i$ be the restriction of $f$ for each $i \in I$.
Since $f$ is proper, $f^{-1}(A_i)$ is compact.
Hence $f_i$ is a closed map.
Hence $f$ is a closed map by Lemma 2.
QED
Definition 2
Let $G$ be a locally compact Hausdorff group.
Let $X$ a locally compact Hausdorff space.
Suppose $G$ acts on $X$ continuously.
Suppose the map $f\colon G\times X \rightarrow X\times X$ defined by $f(s, x) = (x, sx)$ is proper.
Then we say $G$ acts properly on $X$.
Definition 3
Suppose a group $G$ acts on a set $X$.
Let $K, L$ be subsets of $X$.
We denote $P(K, L) = \{s \in G| sK \cap L \neq \emptyset\}$.
Lemma 4
Let $X, Y$ be Hausdorff spaces.
Let $C$ be a compact subset of $X\times Y$.
Then there exists a compact subset $K$(resp. $L$) of $X$(resp. $L$) such that $C \subset K\times L$.
Proof:
Let $K$(resp. $L$) be the image of $C$ by the projection map on the first(resp. the second) factor of $X\times X$.
Then $K$ and $L$ are compact and $C \subset K\times L$.
QED
Lemma 5
Let $f\colon X_1 \rightarrow Y_1$(resp. $g\colon X_2 \rightarrow Y_2$) be a proper map.
Then $f\times g\colon X_1\times X_2 \rightarrow Y_1 \times Y_2$ is proper.
Proof:
Let $C$ be a compact subset of $Y_1\times Y_2$.
By Lemma 4, there exists a compact subset $K$(resp. $L$) of $Y_1$(resp. $Y_2$) such that $C \subset K\times L$.
Since $(f\times g)^{-1}(K\times L) = f^{-1}(K)\times g^{-1}(L)$ is compact, $(f\times g)^{-1}(C)$ is compact.
QED
Lemma 6
Let $X$ be a compact Haudsorff space.
Let $Y$ be a locally compact Hausdorff space.
Then the projection map $\pi\colon X\times Y \rightarrow Y$ is proper.
Proof:
Let $p$ be a one point space.
Then $f\colon X \rightarrow p$ is proper.
Hence, by Lemma 5, $\pi = f\times id_Y\colon X \times Y \rightarrow p\times Y = Y$ is proper.
QED
Lemma 7
Let $G$ be a locally compact Hausdorff group.
Let $X$ a locally compact Hausdorff space.
Suppose $G$ acts on $X$ continuously.
Then $P(K, L)$ is closed for any compact subsets $K, L$ of $X$.
Proof:
Let $f\colon G\times X \rightarrow X$ be the map defined by $f(s, x) = sx$.
Since $f$ is continuous, $f^{-1}(L)$ is closed.
Let $\pi\colon G\times K \rightarrow G$ be the projection.
By Lemma 6, $\pi$ is proper.
Hence, by Lemma 3, $\pi$ is a closed map.
Hence $P(K, L) = \pi(f^{-1}(L))$ is closed.
QED
Lemma 8
Let $G$ be a locally compact Hausdorff group.
Let $X$ a locally compact Hausdorff space.
Suppose $G$ acts on $X$ continuously.
Suppose $P(K, L)$ is compact for any compact subsets $K, L$ of $X$.
Then $G$ acts properly on $X$.
Proof:
Let $h\colon G\times X \rightarrow X\times X$ be the map defined by $h(s, x) = (x, sx)$.
Let $C$ be a compact subset of $X\times X$.
By Lemma 4, there exist compact subsets $K, L$ of $X$ such that $C \subset K\times L$.
Then $h^{-1}(K\times L) \subset P(K, L)\times K$.
Since $P(K, L)\times K$ is compact and $h^{-1}(K\times L)$ is closed, $h^{-1}(K\times L)$ is compact.
Since $C \subset K\times L$, $h^{-1}(C) \subset h^{-1}(K\times L)$.
Since $h^{-1}(C)$ is closed, it is compact.
Hence $h$ is proper.
Hence $G$ acts proplerly on $X$.
QED
Lemma 9
Let $X$ be a topological space.
Let $R$ be an equivalence relation on $X$.
Let $X/R$ be the quotient space.
Let $\pi\colon X \rightarrow X/R$ be the canonical map.
Let $\Gamma = \{(x, y)\in X\times X| x \equiv y$ (mod $R)\}$.
Suppose $\pi$ is an open map and $\Gamma$ is closed.
Then $X/R$ is Hausdorff.
Proof:
Let $\rho := \pi\times \pi\colon X\times X \rightarrow (X/R)\times (X/R)$.
Since $\pi$ is an open map, so is $\rho$.
Since $\rho(\Gamma)$ is the diagonal subset of $(X/R)\times (X/R)$, it suffices to prove that $\rho(\Gamma)$ is closed.
Since $\Gamma = \rho^{-1}(\rho(\Gamma)$, $X\times X - \Gamma = \rho^{-1}((X/R)\times (X/R) - \rho(\Gamma))$.
Since $\rho$ is surjective, $\rho(X\times X - \Gamma) = (X/R)\times (X/R) - \rho(\Gamma)$.
Since $\rho$ is an open map and $X\times X - \Gamma$ is open, $(X/R)\times (X/R) - \rho(\Gamma)$ is open.
Hence $\rho(\Gamma)$ is closed.
QED
Lemma 10
Let $G$ be a locally compact Hausdorff group.
Let $X$ a locally compact Hausdorff space.
Suppose $G$ acts properly on $X$.
Then the orbit space $X/G$ is Hausdorff.
Proof:
Let $\pi\colon X \rightarrow X/G$ be the canonical map.
Let $U$ be an open subset of $X$.
Then $\pi^{-1}(\pi(U)) = GU$.
Since $GU$ is open, $\pi$ is an open map.
By Lemma 9, it suffices to prove that $\Gamma = \{(x, y) \in X\times X| x \equiv y$ (mod $G)\}$ is closed.
Let $h\colon G\times X \rightarrow X\times X$ be the map defined by $h(s, x) = (x, sx)$.
Since $G$ acts properly on $X$, $h$ is proper.
Hence $h$ is closed by Lemma 3.
Hence $\Gamma = h(G\times X)$ is closed.
QED
Proposition
Let $G$ be a compact Hausdorff group.
Let $X$ be a locally compact Hausdorff space.
Suppose $G$ acts continuously on $X$.
Then $G$ acts properly on $X$.
Hence the orbit space $X/G$ is Hausdorff.
Proof:
By Lemma 7, $P(K, L)$ is closed for any compact subsets $K, L$ of $X$.
Since $G$ is compact, $P(K, L)$ is compact.
Hence, by Lemma 8, $G$ acts proplerly on $X$.
Hence, by Lemma 10, $X/G$ is Hausdorff.
QED
Corollary
Let $K$ be the field of real numbers or the field of complex numbers.
Then the projective space $P^n(K)$ over $K$ is Hausdorff.
Proof:
Let $x = (x_1, \dots, x_n) \in K^n$.
We denote $||x|| = (\sum_i x_i \bar x_i)^{\frac{1}{2}}$.
Let $G = \{x \in K|\ |x| = 1\}$.
Then $G$ is a subgroup of the multiplicative group of $K$.
Let $X = \{x \in K^n|\ ||x|| = 1\}$.
Clearly $G$ acts on $X$ continuously.
Since $G$ and $X$ are compact Hausdorff spaces, $X/G$ is Hausdorff by the proposition.
Since $P^n(K) = X/G$, the assertion follows.
QED
Remark 1
The above proposition holds without assuming $X$ is locally Hausdorff, but only assuming it is Hausdorff as the answer to this question shows.
Remark 2
If $G$ and $X$ are both compact Hausdorff, it is easy to prove that $X/G$ is Hausdorff.
Let $\Gamma = \{(x, y) \in X\times X| x \equiv y$ (mod $G)\}$.
Let $h\colon G\times X \rightarrow X\times X$ be the map defined by $h(s, x) = (x, sx)$.
Since $h$ is continuous, $\Gamma = h(G\times X)$ is compact, hence closed.
By Lemma 9, $X/G$ is Hausdorff.