Let $(M,d)$ be a metric space and $f,g:(M,d) \rightarrow \mathbb{R}$ be continuous functions. Show the set $A=\{x \in M: f(x)<g(x)\}$ is an open set in $(M,d)$.
To show the statement we need to take a point $x_0$ and show that there exist a ball around $x_0$ which is contained in $A$. My proof is the following but it does not reasonable to me that I have to have the condition $\epsilon_1>\epsilon_2$.
Let $x_0 \in A \rightarrow f(x_0)<g(x_0)$
Since $f$ is continuous $\forall \epsilon_1>0\,\,\exists\,\,\,\delta_1>0\,\,\,\text{s.t.} \,\,\,d(x,x_0)<\delta_1 \,\,\,\rightarrow |f(x)-f(x_0)|<\epsilon_1$ and $$ -\epsilon_1<f(x)-f(x_0)<\epsilon_1 \,\, \Rightarrow f(x)-f(x_0)<\epsilon_1 \tag{1} $$
And since $g$ is continuous $\forall \epsilon_2>0\,\,\exists\,\,\,\delta_2>0\,\,\,\text{s.t.} \,\,\,d(x,x_0)<\delta_2 \,\,\,\rightarrow |g(x)-g(x_0)|<\epsilon_2$ and $$ -\epsilon_2<g(x)-g(x_0)<\epsilon_2 \,\, \Rightarrow -\epsilon_2 <g(x)-g(x_0)\tag{2} $$
let $\delta=\min (\delta_1,\delta_2)$ so using (1) and (2) $$ f(x)-f(x_0)-\epsilon_2<\epsilon_1 +g(x)-g(x_0) \Rightarrow -\epsilon_2+\epsilon_1< g(x)-f(x) $$ Since $f(x_0)-g(x_0)<0$.
Therefore, if $ \epsilon_1>\epsilon_2 \Rightarrow 0< g(x)-f(x)$ so $d(x,x_0)<\delta$ is contained in $A$.
Could you help me to get rid of this confusion. I understand that because $f,g$ are continuous those statement should be held for all $\epsilon$ but having this condition does not make sense to me.
Since you're not proving continuity, but using continuity, you want to pick a specific $\epsilon$.
Case in point: if $f(x)=0, g(x)=x, x_0=1$ and $\epsilon=2$, then your $\delta$ becomes $2$, and $(-1,3)$ is certainly not a subset of your set. So an arbitrary $\epsilon$ isn't going to do you any good.
I claim that $\epsilon=\frac{g(x_0)-f(x_0)}2$ works. Let $\delta_f>0$ be such that for any $x_1\in M$ with $d(x_0,x_1)<\delta_f$, we have $|f(x_0)-f(x_1))|<\epsilon$, and let $\delta_g$ be defined analogously. Pick $\delta=\min\{\delta_f,\delta_g\}$. Then, for any $x_1\in M$ with $d(x_0,x_1)<\delta$, we have $$ f(x_0)+\frac{g(x_0)-f(x_0)}2=g(x_0)-\frac{g(x_0)-f(x_0)}2\\ f(x_0)+\epsilon=g(x_0)-\epsilon\\ f(x_1)<f(x_0)+\epsilon=g(x_0)-\epsilon<g(x_1) $$ where the first inequality is because $d(x_0,x_1)<\delta_f$ and the second one because $f(x_0,x_1)<\delta_g$.