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 $\varepsilon > 0$ there is $K \in \mathbb{N}$ with $|x_k - z| < \varepsilon$ for all $k \ge K$ (Limits and Cauchy sequences of reals).

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

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

Small rationals: for every real $\eta > 0$ there is a rational $\varepsilon$ with $0 < \varepsilon < \eta$. 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 $1/\eta$, which yields a natural $n \ge 1$ with $1/\eta < n$ and hence $0 < 1/n < \eta$ (Inverses of positives are positive, and reciprocation reverses order).

Order arithmetic in $\mathbb{R}$. Trichotomy, so $u \ne 0$ together with $|u| \ge 0$ and $|u| \ne 0$ forces $|u| > 0$; 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 $\eta/2 = \eta \cdot 2^{-1} > 0$ 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 $\mathbb{N}$).