Proof of the Theorem of sign permanence If I have a metric space with the Euclidean metric $(X,d)$ and a continuous function $f:X\to\mathbb{R}$, assumed that exists a point $x\in X$ such that $f(x)>0$ can I say that exists $\delta>0:f(y)>0, 
 \forall(y)\in B(x,\delta)$, if yes why?
 A: Hint:
That follows from the continuity of $f$.
Prove by contradiction: assume that is not true and see what happens.

Notes.
If the conclusion if not true, then there exists (why? exercise!) a sequence $y_n\in B(x,\frac{1}{n})$ such that $f(y_n)\le 0$. Now think about the limit of $(y_n)$ and $(f(y_n))$.
A: Because $f : X \to \Bbb{R}$ is continuous, if $A$ is any open subset of $\Bbb{R}$, the inverse image $f^{-1}[A]$ is an open subset of $X$. If $f(x) > 0$, put $\alpha = f(x)/2$ (or any positive real with $\alpha < f(x)$) and let $A$ be the open interval $(f(x) - \alpha, f(x) + \alpha)$. Then $Y = f^{-1}[A]$ is an open subset of $X$ such that $f(y) > 0$ for every $y \in Y$. Because $Y$ is an open subset of the metric space $X$ and $x \in Y$, $Y$ contains the open ball $B(x, \delta)$ for all sufficiently small $\delta > 0$.
A: By contradiction if $\forall\delta>0$ there exists $y\in B(x,\delta)$ such that $f(y)=0$ $\Longrightarrow$ then let $B_n:= B(x,1/n)$ so you can pick a $x_n \in B_n$ $\forall n\in \mathbb{N}$; you would have that $x_n \to x$ and, since $f$ is continuous $f(x_n) \to f(x)>0$, wich means $0\to f(x)>0$ which is impossible.

The fact that $x_n\to x$ implies $f(x_n) \to f(x)$ come from the fact that since $F$ is continuous, for all open neighborhod $U$ of $f(x)$, $x\in f^{-1}(U)$ is open in $X$ so it contains a $B_N$ for $N$ sufficiently large, so that $f(x_N)\in U$.
