Quotient space with finite equivalence classes Let ~ be a equivalence relation on $\mathbb{R}$ such that $\mathbb{R}/$~ is finite. 
I need to show that tue quotient space $\mathbb{R}/$~ can't be a Hausdorff space.
My attempt: Supposse that $X=\mathbb{R}/$~$:=\{[x_1],[x_2],\dots,[x_n]\}$ and $X$ is hausdorff. Now, for each $2\leq k\leq n$ I can find a neighborhood $U_{1k}$ of $[x_1]$ such that $[x_k]\notin U_{1k}$ so, the open set $V=\bigcap_{2\leq k\leq n}U_{1k}=\{[x_1]\}$ has a only point. 
With a simillar argument I can find a open set $W$ in $X$ such that $X=U_{1k}\cup W$ and $U_{1k}\cap W=\emptyset$. If $\pi:\mathbb{R}\rightarrow X$ is the canonical map then $\pi^{-1}(U_{1k})\cap\pi^{-1}(W)=\emptyset$ and $\mathbb{R}=\pi^{-1}(U_{1k})\cup\pi^{-1}(W)$. But the last conclution is imposible because $\mathbb{R}$ is connected. This contradiction implies that $X$ is not a Hausdorff space.
Am I right?
 A: Yeah, I believe your proof is correct. But I think you can use $\{[x_1]\}$ instead, to split your space in open sets. 
I don't know if it is clear in your proof that you can really find $W$ such that $U_{1k} \cap W = \emptyset$ and so on. This is how I would do:
We know that $X = \{ [x_1], [x_2], \ldots, [x_n]\}$ is Hausdorff. 
Then, for every $k$ with $1< k \leq n$, there is $U_k \ni x_1$ open such that $x_k \notin U_k$. Notice that
$$\bigcap_{1<k\leq n} U_k = \{[x_1]\}$$ 
therefore, $\{[x_1]\}$ is open.
We can make the same argument to conclude that $\{[x_k]\}$ is open for every $k\leq n$. 
Now, take the open sets $V = \{[x_1]\}$ and $ W = \{[x_2], \ldots, [x_n]\}$ (union of unit sets, thus open). It is clear that $V \cap W = \emptyset$ and $V \cup W = X$. 
By definition of the quotient topology, the projection $\pi: \mathbb{R} \to X$ is continuous, therefore $\pi^{-1}(V)$ and $\pi^{-1}(W)$ are open sets, $\pi^{-1}(V) \cap \pi^{-1}(W) = \emptyset$ and $\pi^{-1}(V) \cup \pi^{-1}(W) = \mathbb{R}$, contradiction with the fact that $\mathbb{R}$ is connected. 
You don't even need Hausdorff to get this result, because this comes from the fact that every finite, $T_1$ space is discrete (https://proofwiki.org/wiki/Finite_T1_Space_is_Discrete) and every discrete space is totally disconnected (https://en.wikipedia.org/wiki/Totally_disconnected_space).

Edit: This does not work if $|\mathbb{R}/\sim| = 1$, since a unitary set is always Hausdorff.
