🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

Suppose for the sake of contradiction that $f$ is uniformly continuous and take $ϵ=1$ therefore we obtain some $r\in {\mathbb{R}}^{+}$ such that for any $x,a\in S$ if $|x-a|<r$ implies that $|f\left(x\right)-f\left(a\right)|<ϵ=1$.

Note that this implies that for any $a,b\in (0,r)$ we have $|f\left(a\right)-f\left(b\right)|<1$ clearly this will lead to nonsense as it blows up around $0$ and so we should be able to find two points whose vertical distance is greater or equal to $1$.

To do this let $x\in (0,r)$, since we know that ${lim}_{x\to 0^{+}}f\left(x\right)=\infty$ therefore by taking $\delta =r,M=f\left(x\right)+1$ we obtain some $a\in (0,r)$ such that $f\left(a\right)>M=f\left(x\right)+1$ therefore we have that $f\left(a\right)-f\left(x\right)>1$ which implies that $${\displaystyle 1>|f\left(a\right)-f\left(x\right)|=|f\left(x\right)-f\left(a\right)|}$$ which is a contradiction, so that $f$ is uniformly continuous.