So I was reading this particular answer https://math.stackexchange.com/a/3113739 and I could not for the life of me prove the statement that was given (as symbolised in the title).
What I tried was that given $f$ and $g$ are continuous, I can choose any $\epsilon_1,\epsilon_2 > 0$ and corresponding $\delta_1,\delta_2$ such that for all $x$ satisfying $|x - c| < \min\{\delta_1,\delta_2\}$ we have \begin{align*} f(c) - \epsilon_1 < f(x) < f(c) + \epsilon_1\\ g(c) - \epsilon_2 < g(x) < g(c) + \epsilon_2 \end{align*} I know that $f(x) > g(x) \iff f(x) - g(x) > 0$, so from here I have \begin{align*} f(x) - g(x) &> (f(c) - \epsilon_1) - (g(c) + \epsilon_2)\\ &= (f(c) - g(c)) - \epsilon_1 - \epsilon_2 \end{align*} Now from here the biggest issue I have is that I don't know what I could possibly set $\epsilon_1$ and $\epsilon_2$ such that the last line above is clearly greater than 0.
Could I please have some help?
First, I'd suggest simplifying your life: by considering $d(x)=f(x)-g(x)$, it is enough to prove the simplified statement that if $d$ is continuous and $d(c)>0$, then there is a neighborhood of $c$ on which $d(x)>0$ as well.
Second, describe the idea of the solution to yourself before the symbols. We know that $d$ is continuous, and so it only changes a little in small neighborhoods. We know that $d(c)>0$. So, we want to pick a neighborhood on which $d(x)$ stays close enough to $d(c)$ that it can't cross below $0$. If $d(x)$ can only differ from $d(c)$ by $\frac{d(c)}{2}$, this will be the case.
To that end, let $\epsilon=\frac{d(c)}{2}$, and note that continuity implies that there exists $\delta>0$ such that $\lvert x-c\rvert<\delta$ implies $\lvert d(x)-d(c)\rvert<\epsilon$.
But that means that $$ -\frac{d(c)}{2}< d(x)-d(c) < \frac{d(c)}{2}, $$ and adding $d(c)$ implies $$ \frac{d(c)}{2}<d(x)<\frac{3d(c)}{2}. $$ But you know that $\frac{d(c)}{2}>0$, so the left inequality proves the result.