Suppose every convergent Sequence has a unique limit point in space $X$, then $X$ is Hausdorff My attempt
Let $a, b \in X$ are two distinct points . we will show that there exist two open sets $G_1$ and $G_2$ such that $a \in G_1$ and $b \in G_2$ and $G_1 \cap G_2 = \phi$ or there  exists $n \in \mathbb N$ such that $S_{1/n}(a) \cap S_{1/n}(b) = \phi$.
Let if possible for each $n \in \mathbb N$ $S_{1/n}(a) \cap S_{1/n}(b) \neq \phi$ , then there exist $x_n \in S_{1/n}(a) \cap S_{1/n}(b)$ for each $n \in \mathbb N$ . So there exist a sequence $\{x_n \} \in X$ suct that converges to two distinct point $a$ and $b$, which is a contradiction.
I would be thankful , if someone check my Solution
 A: The main problem with your proof is that you are trying to prove something that is false.  The following are standard, though not very common, definitions:


*

*A topological space $X$ is called a KC-space is every compact subset of $X$ is closed.

*A topological space $X$ is called a US-space if all convergent sequences have a unique limit point.


Now, it isn't too hard to show that the implications T2 ⇒ KC ⇒ US ⇒ T1 hold. However none of the arrows reverse.
The example provided by Batominovski in a comment to the question

[E]quip $\mathbb R$ with the cocountable topology (i.e. $X \subseteq\mathbb R$ is open iff $\mathbb R \setminus X$ is countable).  Any convergent sequence $\left(x_n\right)_{n\in\mathbb{N}}$ in $\mathbb R$ which converges to $x\in\mathbb R$ cannot converge to another point $y\in\mathbb R$ because $x_n=x$ for all sufficiently large $n\in\mathbb{N}$.

is actually an example of a KC-space which is not Hausdorff. (It is a KC-space because the only compact subsets are finite, which are closed by T1-ness.)
For an example of a US-space which is not a KC-space, consider the space $X = [0,1] \cup \{ \infty \}$ where $[0,1]$ is an open subspace (with the usual topology), and the open neighborhoods of $\infty$ are of the form $\{ \infty \} \cup U$ where $U$ is a dense open subset of $[0,1]$.


*

*This is a US-space essentially because a sequence only converges to $\infty$ if eventually all its points are $\infty$ (and $[0,1]$ is an open set not containing $\infty$).
In more detail, if $(x_n)_n$ is a sequence in $[0,1]$ then by the compactness of $[0,1]$ some subsequence $(x_{n_k})_k$ must converge to a point $x \in [0,1]$. Now it isn't hard to show that $F := \{ x_{n_k} : k \geq 1 \} \cup \{ x \}$ is a closed nowhere dense subset of $[0,1]$, and so $U := \{ \infty \} \cup ( [0,1] \setminus F)$ is an open neighborhood of $\infty$. But then $U$ cannot contain a tail of the sequence $(x_n)_n$, meaning the sequence cannot converge to $\infty$.

*It not a KC-space because $[0,1]$ is a compact subset which is not closed.



The major false move in your attempted proof is the implicit assumption of first-countability, where you are looking at the neighborhoods $S_{1/n} (a)$ and $S_{1/n}(b)$ ($n \in \mathbb N$) for $a,b$, respectively.
But actually, your proof basically gives a correct argument that every first-countable US-space is Hausdorff:

If $X$ is a first-countable space which is not Hausdorff, then there are distinct points $a,b \in X$ which cannot be separated by disjoint open neighborhoods. By first-countability, $a,b$ have descending open neighborhood bases $\{ U_n \}_{n \in \mathbb N}$, $\{ V_n \}_{n \in \mathbb N}$, respectively. By choice of $a,b$, for each $n \in \mathbb N$ we have that $U_n \cap V_n \neq \emptyset$, so pick $x_n \in U_n \cap V_n$. Then $( x_n )_n$ can be shown to be a sequence which converges to both $a$ and $b$. Thus $X$ is not a US-space.

