Spaces in which every sequence of distinct points converges This question is an offshoot of the following recent question: Does there exist a metric on $\mathbb{R}$ such that every sequence converges?
By just adding the word 'distinct' the questions seems to change. An  example of a metric space in which every sequence of distinct points converges is $X=\{0,\frac 1 2, \frac  1 3,...\}$ with the usual metric from $\mathbb R$.  (This space has infinitely many points).
I am wondering if there is any Haudorff topological space with uncountably many points in which every sequence of distinct points converges.
I would also be interested in seeing more interesting examples of metric spces with this property.
 A: It is easy to see that an infinite Hausdorff space has this property, if and only if it has exactly one non-isolated point, and each neighborhood of this point contains all but finitely many points of the space. These spaces are exactly the one-point-compactifications of discrete spaces.
In particular, any metric (or, more generally, first countable) space with this property is homeomorphic to the above example, which has already been indicated in the above comment of Jason DeVito.
Update 
It came to my mind that the uniqueness of the non-isolated point might be less straightforward than I first thought. Therefore, I add a more detailed proof of "$\Rightarrow$":
Say a sequence is distinct, if its elements are pairwise distinct.
Let $X$ be an infinite Hausdorff space such that each distinct sequence in $X$ converges.
Then $X$ has exactly one non-isolated point, and each neighborhood of this point contains all but finitely many points of the space. In particular, $X$ is compact and therefore the one-point compactification of its discrete subspace of isolated points.
PROOF.
Choose a distinct sequence
$(x_n)_{n \in \mathbb N}$ in $X$.
Let $x$ be its limit point. Of course, $x$ is not isolated.
Now assume $x \neq y$ is another non-isolated point of $X$. By T2, there are disjoint, open neighborhoods $U, V$ of $x, y$, respectively.
Pick $m \in \mathbb N$ such that $x_n \in U$ for all $n \ge m$.
Since $X$ is T2 and $y$ is not isolated, $V$ is infinite. Hence, we can choose a distinct sequence
$(y_n)_{n \in \mathbb N} \subset V$.
The sequence
$(x_m, y_m, x_{m+1}, y_{m+1}, \dots)$ is distinct, hence it has a limit point $y$.
Since also $(x_m, x_{m+1}, \dots) \rightarrow y$, again by T2, we have $x = y$. Contradicition!
Hence, $x$ is the unique non-isolated point of $X$.
Now let $W$ be an arbitrary neighborhood of $x$. Assume $X \setminus W$ is infinite and choose a distinct sequence
$(w_n)_{n \in \mathbb N} \subset X \setminus W$.
Since this sequence converges to a (necessarily non-isolated) point, it converges to $x$. Contradiction! Hence, $X \setminus W$ is finite.
