Show $(\mathbb{R}, \tau_{co-countable})$ is not Hausdorff but every sequence converge to at most one point 
Given $(\mathbb{R}, \tau_{co-countable})$, show that it is not
  Hausdorff but every sequence converges to at most one point.


1. If  $(\mathbb{R}, \tau_{co-countable})$ is not Hausdorff, then $\exists x,y \in \mathbb{R}$ such that exists $U,V \in \tau_{co-countable}$, $x \in U, y \in V$ and $U \cap V \neq \varnothing$ 
(I hope this is the correct negation)
Then let $x \in U = \mathbb{R}\backslash\{\mathbb{R}\backslash U\}$, where $U$ is countable. Likewise $y \in \mathbb{R}\backslash\{\mathbb{R}\backslash V\}$, $V$ countable. 
But here is what I encounter a problem. Set $U = \{x\}$ and $V = \{y\}$, wouldn't this be examples of co-countable open sets that separates the two points?

2. Let $(x_n)$ be a sequence in $\mathbb{R}$. Suppose that $x_n \to x$, want to show that $x_n \not \to y, x \neq y$.
Recall that $x_n \to x$ iff $\forall U \in \tau_{co-countable}, x \in U, \exists N \in \mathbb{N}$ s.t. $x_n \in U, \forall n \geq N$
Then $x \in U = \mathbb{R}\backslash\{\mathbb{R}\backslash U\}$, $U$ is countable. In particular, let $U = \{x\}$, then the only sequence that converges are the ones that are eventually constant. Since $x \not \in \{y\}$, therefore $x_n \not \to y$.
 A: First, your negation of ‘$X$ is Hausdorff’ is not correct. To show that a space is not Hausdorff, you must show that there are distinct points $x$ and $y$ in the space such that if $U$ is any open nbhd of $x$, and $V$ is any open nbhd of $y$, then $U\cap V\ne\varnothing$. That is, you must show that $X$ contains two distinct points that do not have disjoint open nbhds.
The next problem is that you’ve misunderstood the co-countable topology. If $U$ is an open set containing $x$, then $\Bbb R\setminus U$ is countable, not $U$. The open sets are the complements of the countable sets, not the countable sets themselves. The set $\{x\}$ is countable, so $\Bbb R\setminus\{x\}$ is open; but $\Bbb R\setminus\{x\}$ is not countable, so $\{x\}=\Bbb R\setminus(\Bbb R\setminus\{x\})$ is not open.
Let $x,y\in\Bbb R$ with $x\ne y$. Suppose that $U$ is an open nbhd of $x$, and $V$ is an open nbhd of $y$. Let $C=\Bbb R\setminus U$ and $D=\Bbb R\setminus V$; by definition $C$ and $D$ are countable subsets of $\Bbb R$. Now
$$U\cap V=(\Bbb R\setminus C)\cap(\Bbb R\setminus D)=\Bbb R\setminus(C\cup D)\;.$$
$\Bbb R$ is uncountable, and $C\cup D$ is countable (why?), so $\Bbb R\setminus(C\cup D)\ne\varnothing$, and therefore $U\cap V\ne\varnothing$. We’ve just proved that no matter what open nbhds of $x$ and $y$ we choose, they are not disjoint: their intersection is not empty. Thus, $x$ and $y$ cannot be separated by disjoint open sets, and $\Bbb R$ is not Hausdorff with the co-countable topology.
Your misunderstanding of the co-countable topology also invalidates your answer to the second question. I’d like to give you another chance at answering that one correctly after you get the confusion about the topology sorted out, so I’ll just give a hint:


*

*Show that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ in the co-countable topology if and only if there is an $N\in\Bbb N$ such that $x_n=x$ for all $n\ge N$.


From that result you can easily deduce that the limit, when it exists, is unique.
