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

Note that this implies that for any $$ a,b\in <!--\; \in \; -->(0,r)$$ 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 $$ x\in <!--\; \in \; -->(0,r)$$, since we know that $$ \underset{x\to <!--\; \to \; -->0^{+}}{lim}f\left(x\right)=\mathrm{\infty <!--\; \infty \; -->}$$ therefore by taking $$ \mathrm{\delta <!--\; \delta \; -->}=r,M=f\left(x\right)+1$$ we obtain some $$ a\in <!--\; \in \; -->(0,r)$$ 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 $\[ 1><!--\; >\; -->|f\left(a\right)-<!--\; -\; -->f\left(x\right)|=|f\left(x\right)-<!--\; -\; -->f\left(a\right)| \]$ which is a contradiction, so that $$ f$$ is uniformly continuous.