Hausdorff property of $\mathbb{RP}^n$ from unusual definition Rather than defining the topology on $\mathbb{RP}^n$ as the quotient $(\mathbb{R}^{n+1}\backslash\{0\})/$~ or $S^n/$~ in the usual way, suppose you use these equivalence relations simply to define a set. Then give that set charts $(\varphi_i,U_i)$ where $U_i = \{[x_o,\dots,x_m]:x_i\not=0\}$ and $\varphi_i[x_0,\dots,x_n]=(x_0,\dots,x_n)$ and then induce the topology on $\mathbb{RP}^n$ from these charts. 
This clearly satisfies all the properties of being a manifold with differentiable structure determined by the charts $(\varphi_i,U_i)$, except that I'm not sure how to show that it is Hausdorff without resorting to showing that it's is homeomorphic to the "usual" $\mathbb{RP}^n$. Certainly I can separate points that lie within the same chart, but what about if they don't?
 A: Let $a\in U_0$ and $b\notin U_0$. Write $a = [1, a_1, \cdots, a_n]$. WLOG let $b\in U_1$. Then $b = [0, 1, b_2, \cdots, b_n]$. 
Let $\epsilon >0$ and consider 
$$B_b(\epsilon)=\{[c_1,1, c_2, \cdots c_n]: (c_1-0)^2 + (c_2 - b_2)^2 + \cdots +(c_n-b_n)^2 < \epsilon\} \subset U_1$$
This is open using your defintion. Now we want to find $\epsilon>0$ small so that $B_a(\epsilon) \cap B_b(\epsilon) = \emptyset$, where 
$$B_a(\epsilon) = \{[1, d_1, \cdots, d_n]: (d_1 - a_1)^2 + \cdots +(d_n-a_n)^2 <\epsilon\} \subset U_0.$$
Note that if $c_1 = 0$, then $[c_1 , 1, c_2, \cdots, c_n] \notin B_a(\epsilon)$. If $c_1 \neq 0$, then 
$$[c_1, 1, c_2, \cdots , c_n] = [1, 1/c_1, c_2/c_1, \cdots c_n/c_1]$$
Then choose $\epsilon$ so that $|a_1| + \epsilon  < 1/\epsilon $. As $|c_1|<\epsilon$, we have $1/|c_1| > 1/\epsilon > |a_1| + \epsilon$. Thus 
$$ | 1/c_1 - a_1| \geq |1/c_1| - |a_1| > \epsilon$$
This shows that $||(1/c_1, c_2/c_1, \cdots c_n/c_1) - (a_1, \cdots a_n)|| > \epsilon$ and so $[c_1, 1, c_2, \cdots c_n]$ is not in $B_a(\epsilon)$. Thus the space if really Hausdorff. 
