Let $K$ be an ordered field. Then $K$ contains a unique copy of the ordered field $\mathbb{Q}$. Also, we can define the absolute value $|\cdot|$ and sequence convergence in exactly the same way as in $\mathbb{R}$. In particular, a sequence $(a_n)\subset K$ is said to converge to $l\in K$ if $$ \forall\epsilon\in K(\epsilon>0\to\exists N\in\mathbb{N}(\forall n\in\mathbb{N}(n\geq N\to|a_n-l|<\epsilon))) $$
Question: In general, will $l$ be unique? If not, what is wrong with the usual proof given below?
I'm asking since it was remarked here that it will not in general. The example of the hyperreals in which every other point infinitesimally close to $l$ will also be a limit of $(a_n)$ was given.
Proof (possibly flawed): Suppose $l,l'$ are two limits of $(a_n)$. WLOG, suppose that $l'<l$. Let $\epsilon:=\frac{l'-l}{2}>0$, so that there exists $N_1\in\mathbb{N}$ such that if $n\geq n_1$ then $$ |a_n-l|<\frac{l'-l}{2} $$ Also, there exists $N_2\in\mathbb{N}$ such that if $n\geq N_2$ then $$ |a_n-l'|<\frac{l'-l}{2} $$ If $N:=\max\{N_1,N_2\}$, then $$ a_N-l<\frac{l'-l}{2}\text{ and }-\frac{l'-l}{2}<a_N-l' $$ so $$ \frac{l'+l}{2}<a_N<\frac{l'+l}{2} $$ which is impossible.
The proof of uniqueness works fine. The problem (if one thinks of it as a problem) is that for example the sequence $(1/n)$ need not have a limit.