If $f$ is continuous on $(0,1)$ and $\lim_{x \to 0^+}f(x) = +\infty$, show that $f$ is not uniformly continuous 
If $f$ is continuous on $(0,1)$ and $\lim_\limits{x \to 0^+}f(x) = +\infty$, show that $f$ is not uniformly continuous.

Edit: There is a similar question that has been answered, but this one discusses the reasoning behind certain details in more depth.
Attempt
Suppose $f$ is uniformly continuous. Then $\forall\epsilon > 0$ there exists $\delta > 0$ s.t for all $x,y \in S \subset \mathbb{R}$, if $\lvert x-y\rvert < \delta$ then $\lvert f(x) - f(y)\rvert < \epsilon$.
But $\lim\limits_{x \to 0^+}f(x) = +\infty \implies \lvert f(x) -f(y)\rvert = +\infty > \epsilon$.
Comment: I know that this is not formally right due to how I'm using infinity, but I can't figure out how to formalize what I'm trying to say. I understand what is going on with the function and why it can't be uniformly continuous, but expressing it is becoming difficult.

Let's suppose it were uniformly continuous. Pick $\varepsilon > 0$ and the corresponding $\delta > 0$ as in the definition. Choose also an $x$ with $0 < x < \delta$. By uniform continuity, for any $y$ with $0 < y < x$, $|f(x) - f(y)| < \epsilon$. But then if we take the limit as $y$ approaches $0$ from the right, then $$\lim_{y\to0^+}|f(y)| \le |f(x)| + \lim_{y\to0^+}|f(y) - f(x)| < |f(x)| + \epsilon.$$
This contradicts the fact that $\lim_{x\to0^+}f(x) = +\infty$, so $f$ cannot be uniformly continuous.
