I would like to know if my proof about the uniqueness of the limit of a function is correct. I have already seen a proof about it which uses a different contradiction that the one I use.
Proof:
Suppose $m, l$ are both limits of $f\left(x\right)$ for $x \to a$, with $m \neq l$. This means that, $\forall \varepsilon > 0$,
$$\exists \delta_1 > 0, \delta_2 > 0 \textrm{ s.t. } |x-a|< \delta_1 \implies |f\left(x\right)-l| < \varepsilon \textrm{ and } |x-a| < \delta_2 \implies |f\left(x\right) - m| < \varepsilon$$
By picking $\delta = \min\left(\delta_1, \delta_2\right)$ we get that
$$|x-a| < \delta \implies |f\left(x\right) - l| < \varepsilon \textrm{ and } |f\left(x\right) - m| < \varepsilon$$
By picking $\varepsilon = |l-m|/2$ we have
$$|f\left(x\right) - l| < |l-m| / 2 = |l - f\left(x\right) + f\left(x\right) - m|/2 \leq |f\left(x\right)-l|/2 + |f\left(x\right)-m|/2$$
So we have
$$|f\left(x\right) - l| < |f\left(x\right) - m|$$
By the completeness of $\mathbb{R}$ we have $\varepsilon_R$ s.t. $|f\left(x\right) - l| < \varepsilon_r < |f\left(x\right) - m|$, which contradicts that $m$ is a limit of $f\left(x\right)$ for $x \to a$.
Update:
The contradiction is caused by the fact that for $\varepsilon_R$, all $\forall \delta > 0, |x-a| < \delta$ doesn't imply $|f\left(x\right) - m| < \varepsilon_R$.
The reason is that in our proof we picked $\delta = \min\left(\delta_1, \delta_2\right)$, but in fact we could choose every $\delta$ s.t. $0 < \delta < \min\left(\delta_1, \delta_2\right)$ and our proof would still hold. We don't care for bigger $\delta$s because it would mean that $|f\left(x\right)-m|$ is not a limit for $\varepsilon = |l-m|/2$
Update 2:
One shall keep in mind that in the steps of our proof we showed that
$$\forall \delta \textrm{ s.t. }0 < \delta < \min\left(\delta_1, \delta_2\right), |x-a| < \delta \implies |f\left(x\right)-l| < \varepsilon_r < |f\left(x\right)-m|$$
$\varepsilon_r$ isn't a fixed value: its value is picked based on $x$ as long as $|x - a| < \delta$. We know about the existence of $\varepsilon_r$ by the completeness of the reals, but we don't know anything about $|f\left(x\right)-l| - |f\left(x\right)-m|$ if not that such difference is $> 0$. One could also think about this as the fact that $f\left(x\right)$ approximates $l$ better than how it approximates $m$