A sequence has at most one limit

Given: A sequence $(x_k)$ of reals and reals $x,y$ such that $(x_k)$ converges to $x$ and $(x_k)$ converges to $y$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

Convergence: $(x_k)$ converges to $z$ when for every rational $\epsilon >0$ there is $K\in \mathbb{N}$ with $\mathrm{\mid }{x}_k-z\mathrm{\mid }<\epsilon$ for all $k\ge K$ (Limits and Cauchy sequences of reals).

Triangle inequality: $\mathrm{\mid }a+b\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }$ in any ordered field, in particular in $\mathbb{R}$ (The triangle inequality, Complete ordered field (least-upper-bound property)).

Absolute value: $\mathrm{\mid }u\mathrm{\mid }\ge 0$, and $\mathrm{\mid }u\mathrm{\mid }=0$ if and only if $u=0$, and $\mathrm{\mid }-u\mathrm{\mid }=\mathrm{\mid }u\mathrm{\mid }$ (Basic properties of the absolute value).

Small rationals: for every real $\eta >0$ there is a rational $\epsilon$ with $0<\epsilon <\eta$. Either route gives this: density of $\mathbb{Q}$ in $\mathbb{R}$ (The rationals embed densely in the reals) applied to the pair $0<\eta$; or the Archimedean property (Every complete ordered field is Archimedean) applied to $1\mathrm{/}\eta$, which yields a natural $n\ge 1$ with $1\mathrm{/}\eta <n$ and hence $0<1\mathrm{/}n<\eta$ (Inverses of positives are positive, and reciprocation reverses order).

Order arithmetic in $\mathbb{R}$. Trichotomy, so $u\mathrm{\ne }0$ together with $\mathrm{\mid }u\mathrm{\mid }\ge 0$ and $\mathrm{\mid }u\mathrm{\mid }\mathrm{\ne }0$ forces $\mathrm{\mid }u\mathrm{\mid }>0$; transitivity and irreflexivity of $<$; and, since $u\le v$ means $u<v$ or $u=v$, the mixed form $u\le v<w\Rightarrow u<w$ (Complete ordered field (least-upper-bound property), Ordered field). Adding two strict inequalities: $a<b$ and $c<d$ give $a+c<b+d$ (Order is preserved by adding a constant and by adding inequalities). Multiplying by a positive: for $c>0$, $a<b$ gives $ac<bc$ (Sign rules for products and monotonicity of multiplication). Halving a positive: $1>0$ (The multiplicative identity is positive), so $2:=1+1>0$ because the positives are closed under addition (Ordered field), hence $2^{-1}>0$ (Inverses of positives are positive, and reciprocation reverses order) and $\eta \mathrm{/}2=\eta \cdot 2^{-1}>0$ whenever $\eta >0$ (Sign rules for products and monotonicity of multiplication).

The order on $\mathbb{N}$ is total, so any two indices $K_1,K_2$ admit an index $k$ with $k\ge K_1$ and $k\ge K_2$ ($\le$ is a linear order on $\mathbb{N}$).