Same sign in an open neighborhood. Prove that if a function $f$ is continuous at $x_0$ with $f(x_0)>0$, then there is an open neighborhood I around that $x_0$ (which means $I=(a,b)$ with $x_0$ belonging in $(a,b)$), such that $f(x)>0$ everywhere in $I$.
My thoughts:
1. I was thinking that maybe the method of bisection could help. Like taking a space for example $C=(x_1,x_2)$ and checking if $f(x_1)>0$ and $f(x_2)>0$. If both of the are positive then that's our $I=(a,b)$. I really don't know if that's by any chance even relevant!
2. I thought to prove that the function is Lipschitz continuous and from the Theorem that says if a function is Lipschitz continuous in an $I$, then it's continuous $I$.
Note: The exercise just mentions that the function is continuous at $x_0$. 
 A: "$f$ continuous at $x_0$" means that for every open set $V$ with $f(x_0)\in V$ there is an open set $U$ with $x_0\in U$ and $U\subseteq f^{-1}(V)$.
I will preassume that you are dealing with a function $f:\mathbb R\to\mathbb R$ continuous w.r.t. the usual topology on $\mathbb R$ (domain and codomain).
We have $f(x_0)\in(0,\infty)$ and $(0,\infty)$ is a an open set. Then according to the mentioned definition we can find an open $U\subseteq\mathbb R$ with $x_0\in U$ and $U\subseteq f^{-1}((0,\infty))$. Note that $U\subseteq f^{-1}((0,\infty))$ means exactly that $f(x)>0$ for each $x\in U$.
The fact that $U$ is open combined with the fact that $x_0\in U$ implies the existence of elements $a,b\in \mathbb R$ satisfying $a<x_0<b$ and $(a,b)\subseteq U$. Then $x_0\in(a,b)\subseteq U\subseteq f^{-1}((0,\infty))$ so that $f(x)>0$ for each $x\in(a,b)$.
A: You only need to apply the definition for continuity. For every $\epsilon>0$ theres a $\delta$ such that $f(x_0)-\epsilon<f(x_0+h)<f(x_0)+\epsilon$ whenever $|h|<\delta$.
Choose $\epsilon = f(x_0)$ then you have that $f(x_0+h)>0$ if $|h|<\delta$. That is $f$ is positive on $I=(x_0-\delta, x_0+\delta)$.
