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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.

Suppose $x_n rightarrow x$, then if $x neq y$ there exists neighbourhoods $U,V$ of $x,y$ respectively that are disjoint by Hausdorfness. Next by 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.

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 $x,yin X$ with $xneq y$ then we can find $U,V$ open sets with $xin U, yin V$ s.t. $Ucap V=emptyset$.

Okay, now suppose that $x_nto x,y$ and we want to prove that $x=y$. Suppose that this was not true, then we can find $U,V$ as above; but $x_nto x$ means by definition that for any open set $A$ with $xin A$ $(x_n)$ is contained in $A$ from some index and on. Do this for $U,V$ and you immediately have a contradiction, since they are disjoint.

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.