$|a-b|<\varepsilon\implies a=b$ for uniqueness of limit? Consider the following statement:
Let $a,b\in\mathbb{R}$ be any two real numbers. Then we have that
$$\forall\varepsilon>0:|a-b|<\varepsilon\implies a=b.$$
Here is my attempt to prove it:
Proof. Assume that $a\neq b$. Without loss of generality we may assume that $a<b$ (else we change the names of $a$ and $b$).  $|a-b|=|b-a|=b-a>0$. Let $\varepsilon:=\frac{b-a}{2}>0$. By assumption we have that
$$b-a=|a-b|<\varepsilon=\frac{b-a}{2}\implies 1<\frac{1}{2}$$
which is a contradiction. Therefore we must have that $a=b$.
q.e.d.
Is my proof correct? Can I use this statement to prove that the limit of a sequence $(a_n)_n$ is unique via assuming it has two limits $a,b$ and then concluding that for any $\varepsilon>0$
$$|a-b|\leq|a-a_n|+|a_n-b|<\varepsilon$$
for $n$ suffeciently large? (Because the statement then states that $a=b\implies$ uniqueness of limit).
Thank You
EDIT: I corrected the last line by replacing $=$ by $\leq$ since I used the triangle inequality.
 A: Yes, your answer is correct. Only one little mistake. Make sure that you use the triangle inequality in your last equation,
$$
|a-b| \leq |a- a_n| + |a_n - b| < 2 \epsilon
$$
and use the definition of sequence convergence:
For any $\epsilon > 0$ there is a $N \in \mathbb{N}$, such that for all $n \geq N$ one has
$$
|a- a_n| < \epsilon
$$
A: In fact, according to your reasoning, if we suppose WLOG that $b > a$, we can also proceed as follows:
\begin{align*}
\varepsilon = k(b-a) & \Rightarrow |a - b| = b - a < k(b-a)\\\\
& \Rightarrow (1 - k)(b - a) < 0\\\\
& \Rightarrow b - a < 0\\\\
& \Rightarrow b < a
\end{align*}
whence we get a contradiction (if we assume that $k\in(0,1)$).
If you want to prove that limits are unique, you can proceed more generally as it follows.
Suppose we are given a metric space $(X,d_{X})$ and a sequence $x_{n}\in X$ which converges to $L$.
Let us also assume that $x_{n}$ converges to $M$. We shall prove that $L = M$.
Indeed, if $L\neq M$, according to the definition of convergence in metric spaces, precisely:
\begin{align*}
(\forall\varepsilon > 0)(\exists n_{\varepsilon}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{\varepsilon}\Rightarrow d_{X}(x_{n},L) < \varepsilon)
\end{align*}
We can take $\varepsilon = kd_{X}(L,M)$, where $k\in(0,1)$.
Consequently, there are $n^{1}_{\varepsilon}\in\mathbb{N}$ and $n^{2}_{\varepsilon}\in\mathbb{N}$ such that
\begin{align*}
\begin{cases}
n\geq n^{1}_{\varepsilon} \Rightarrow d_{X}(x_{n},L) < \varepsilon/2\\\\
n\geq n^{2}_{\varepsilon} \Rightarrow d_{X}(x_{n},M) < \varepsilon/2
\end{cases}
\end{align*}
we take $n_{\varepsilon} = \max\{n^{1}_{\varepsilon},n^{2}_{\varepsilon}\}$, we conclude that
\begin{align*}
n\geq n_{\varepsilon} & \Rightarrow d_{X}(L,M) \leq d_{X}(x_{n},L) + d_{X}(x_{n},M) \leq \varepsilon = kd_{X}(L,M)
\end{align*}
whence we conclude that $L = M$, a contradiction.
Hopefully this helps!
A: Your answer is fine, but there is no reason to consider the two cases $a<b$ and $a>b,$ even if you only eliminate one or the using a WLOG argument.
Lemma: For real $x\geq 0$ and $x<\varepsilon,$ for all $\varepsilon>0,$ then $x=0.$
Proof: Esssentially the same. If $x\neq 0,$ then $x>0,$ and we can let $\varepsilon =x/2>0.$

Then your result follows with $x=|a-b|,$ since it is easy to show $|a-b|=0$ iff $a=b.$

This argument works with $a,b$ complex numbers, too, since $|a-b|\geq 0,$ even though we don’t have a notion of order in complex numbers.
It also works in any metric space, with $|a-b|$ replaced by the metric $d(a,b)$ of the metric space.
