Proof that limit in a Hausdorff space is unique This proof is likely quite trivial, but I was hoping someone could look it over regardless. There is one particular step I am confused on.
Theorem. The limit of a convergent sequence in a Hausdorff space is unique. 
Proof. Let $a_n$ be a convergent sequence in a Hausdorf space. Suppose, for a contradiction, that it converges to two different points, $x$ and $y$. Thus, it follows from converges to $x$ that
\begin{align*}
\forall \epsilon > 0, \exists N, \forall n > N, |a_n - x | < \epsilon,
\end{align*}
which is otherwise stated that for all $n > N$, elements of the sequence lie in some open ball around $x$ with radius $\epsilon$. 
From convergence to $y$, it follows that
\begin{align*}
\forall \epsilon > 0, \exists N, \forall n > N, |a_n - y| < \epsilon,
\end{align*}
otherwise stated that for all $n > N$, elements of the sequence lie in an open ball around $x$ with radius $\epsilon$. 
Here is where my confusion comes in. From here, I know I need to draw on the definition of Hausdorff space. These are distinct points, and so there exist open sets around them containing the points, $x$ and $y$, where these sets are disjoint. This does not imply that every open set containing these points is disjoint. So, it seems that I need to say something to the effect that the definition of convergence allows me to create an open ball (I am using this interchangeable with open set, which I hope isn't incorrect; please correct me, if so) of any radius I want around 
the points, and thus it clearly captures all such open sets. Thus, I can pick two separate $N$'s for each of these sets to form open balls of radius $\epsilon_1$ and $\epsilon_2$ around these points such that the sets are disjoint, which I know I can do via the definition of Hausdorff space. Since this would be true for an infinite number of $n$ past some arbitrary point $N$, it would not be possible to get "back inside" the opening ball around the other point. That's clearly a contradiction to the definition of convergence. Thus, if $a_n$ converges to $x$, it cannot converge to $y$, and if it converges to $y$, it cannot converge to $x$, so there is only a single possible limit point, which is unique. 
How does this argument sound? Is there a better way to state it, or have I made an errors in logic?
Thanks in advance. 
 A: Suppose $x_n \rightarrow x$, then if $x \neq y$ there exists neighbourhoods $U,V$ of $x,y$ respectively that are disjoint by Hausdorfness. By the definition of convergence, $U$ must contain all but finitely many of the $x_n$, and so $V$ cannot, and so $x_n$ cannot converge to $y$, because $V$ is a neighbourhood of $y$ that does not contain all but finitely many of the $x_n$. I think you should use this definition of convergence of a sequence in a topological space, given in Munkres Topology.
A: You are reasoning as if you're in the reals, and in the reals your argument isn't valid either as you wrote it. Just use definitions and the proof writes itself:
Suppose, for a contradiction, that for some sequence $(x_n)$ from $X$ we have $x,y \in X$ with $(x_n) \to x$ and $(x_n) \to y$ and $x \neq y$.
By the Hausdorff property, there are disjoint open sets $U$ and $V$ of $X$ such that $x \in U$ and $y \in V$.
As $x_n \to x$, and $U$ is an open neighbourhood of $x$, there is a $N_0 \in \mathbb{N}$, such that for all $n \ge N_0$ we have $x_n \in U$. $(1)$
Also, as $x_n \to y$, and $V$ is an open neighbourhood of $y$, there is a $N_1 \in \mathbb{N}$, such that for all $n \ge N_1$ we have $x_n \in V$. $(2)$
Now let $m= \max(N_0, N_1)$, then $m \ge N_0$ so $x_m \in U$ by $(1)$ and also
$m \ge N_1$ so by (2) we have $x_m \in V$.
But then $x_m \in U \cap V$ contradicts the disjointness of $U$ and $V$.
This shows that all convergent sequences have a unique limit in a Hausdorff space. 
A: I would advise you to forget about radii when studying general topology: neighborhoods are a way more general notion; they do not come in with a radius.
Anyway, remember the definition of a Hausdorff space and that's all you're gonna need: If $x,y\in X$ with $x\neq y$ then we can find $U,V$ open sets with $x\in U, y\in V$ s.t. $U\cap V=\emptyset$.
Okay, now suppose that $x_n\to x,y$ and we want to prove that $x=y$. Suppose that this was not true, then we can find $U,V$ as above; but $x_n\to x$ means by definition that for any open set $A$ with $x\in A$ $(x_n)$ is contained in $A$ from some index and on. Do this for $U,V$ and you immediately have a contradiction, since they are disjoint.
