Proof of Uniqueness of Limits in a Metric Space

Let $\mathbb{N}$ be the natural numbers with the order $\le$ and addition $+$. The values of $d$ are real numbers, whose order $\le$ is that of an ordered field, with multiplicative identity $1$ and multiplicative inverses of the underlying field; for real numbers $s,t$ write $s<t$ to mean $s\le t$ and $s\ne t$, and set $2=1+1$.

Suppose, for contradiction, that $x\ne y$. By condition 2 in the definition of a metric, $d(x,y)=0$ would force $x=y$, so $d(x,y)\ne 0$; condition 1 gives $0\le d(x,y)$, and therefore $0<d(x,y)$.

Put $\varepsilon=d(x,y)\cdot 2^{-1}$. By statement 8 of Elementary Order Arithmetic in an Ordered Field the inverse $2^{-1}$ exists, $0<\varepsilon$, and

Since $(x_m)_{m\in\mathbb{N}}$ converges to $x$, the definition of convergence gives $N_1\in\mathbb{N}$ with $d(x_m,x)<\varepsilon$ for every $m\in\mathbb{N}$ with $N_1\le m$; since it converges to $y$, there is likewise $N_2\in\mathbb{N}$ with $d(x_m,y)<\varepsilon$ for every $m$ with $N_2\le m$.

Put $N=N_1+N_2$. Statement 6 of Properties of the Order on the Natural Numbers gives $N_1<N_1+N_2$ and $N_2<N_2+N_1$, and addition on $\mathbb{N}$ is commutative by statement 4 of Arithmetic of Addition on the Natural Numbers, so $N_2+N_1=N$; statement 1 of Properties of the Order on the Natural Numbers then gives $N_1\le N$ and $N_2\le N$. Hence

Condition 4 in the definition of a metric gives $d(x,y)\le d(x,x_N)+d(x_N,y)$, and condition 3 gives $d(x,x_N)=d(x_N,x)$, so

Applying statement 3 of Elementary Order Arithmetic in an Ordered Field to the strict inequality $d(x_N,x)<\varepsilon$ and the inequality $d(x_N,y)\le\varepsilon$, which holds because $d(x_N,y)<\varepsilon$, gives

Mixed transitivity, statement 2 of Elementary Order Arithmetic in an Ordered Field, now gives $d(x,y)<d(x,y)$, so in particular $d(x,y)\ne d(x,y)$, which is false.

This contradiction shows that $x=y$.