Proof that limits are unique

It may be obvious, but in math we like to prove things.

Let $c$ and $d$ be real numbers. Then $|c-d|\ge0$, so $|c-d|=0$ or $|c-d|>0$.

Assume, with an eye toward contradiction, that $|c-d|>0$,

and for all $\epsilon>0$ we have $|c-d|<\epsilon.$

But taking $\epsilon=|c-d|$, then we would have $\epsilon<\epsilon$, which is a contradiction.

Thus, our assumption that $|c-d|>0$ yielded a contradiction, and it must be that $|c-d|=0$.

Thus, $c=d$.

Here it is the proof in the context of metric spaces for the sake of curiosity.

Proposition

Let $f:(X,d_{X})\to(Y,d_{Y})$ be a function between metric spaces, $x_{0}\in X$ such that $x_{0}$ is an adherent point of $E\subseteq X$ and $L\in Y$. If $f$ converges to $L$ as $x$ approaches $x_{0}$ along $E$, then such limit is unique.

Proof

Let us assume that $f$ converges to $L$ and $L'$ as $x$ approaches $x_{0}$ along $E$, and $L \neq L'$.

Then we can take $\varepsilon = d_{Y}(L,L') > 0$.

For such value of $\varepsilon$, there corresponds $\delta^{1}_{\varepsilon} > 0$ and $\delta^{2}_{\varepsilon} > 0$ such that for every $x\in E$ one has that \begin{align*} \begin{cases} \displaystyle d_{X}(x,x_{0}) \leq \delta^{1}_{\varepsilon} \Rightarrow d_{Y}(f(x),L) \leq \frac{\varepsilon}{3} = \frac{d_{Y}(L,L')}{3}\\\\ \displaystyle d_{X}(x,x_{0}) \leq \delta^{2}_{\varepsilon} \Rightarrow d_{Y}(f(x),L') \leq \frac{\varepsilon}{3} = \frac{d_{Y}(L,L')}{3} \end{cases} \end{align*}

Thus, for such value of $\varepsilon > 0$, there corresponds $\delta_{\varepsilon} = \min\{\delta^{1}_{\varepsilon},\delta^{2}_{\varepsilon}\}$ s.t. for every $x\in E$ one has that \begin{align*} d_{X}(x,x_{0}) \leq \delta_{\varepsilon} \Rightarrow d_{Y}(L,L') & \leq d_{Y}(f(x),L) + d_{Y}(f(x),L')\\\\ & \leq \frac{d_{Y}(L,L')}{3} + \frac{d_{Y}(L,L')}{3} = \frac{2d_{Y}(L,L')}{3}\\\\ & \Rightarrow d_{Y}(L,L') \leq 0 \Rightarrow L = L' \end{align*} which contradicts our assumption.

Thus, if the limit exists, it is unique.

Hopefully this contributes!

This is actually not provable without really getting deep into the definition/construction of $\mathbb R$. We need the fact that if $a \neq b$, then $|a-b|>0$, and proving that, rather than taking it as an axiom, requires rather complicated mathematics.