Trying to find a counter example If $f$ is bounded and continuous on $R^n$ to $R$ and if $f(x_0) >0$. 
1) Show that $f$ is strictly positive on some neighborhood of $x_0$.
2) Does the same conclusion hold if $f$ is merely continuous at $x_0$? 

1) It's pretty simple really. Since $f$ is continuous, then by definition we have that if $V$ is a neighborhood of $f(x_0)$ such that $V = \{y \in R(f) : y > 0 \} $ then there is a corresponding neighborhood $U \subset D(f)$ of $x_0$ such that every elements in $U$ get map inside $V$. Thus $f$ is strictly positive on the neighborhood $U$ of $x$. \
2) This is the part that kinda bother me. It seems like it's true without requiring the fact that the function $f$ to be bounded. But somehow I feel like there is an importance reason why the author put the bounded condition there. Any advice??? 
 A: Did you use the fact that $f$ is bounded in your answer to $(1)$? No. Therefore it doesn't matter whether $f$ is bounded or not.
A: Although the OP sailed through $\text{(1)}$ with 

$\quad$ "It's pretty simple really."

I am 'bothered' by their definition of $V$ (see Arnaud Mortier's comment).
The $R(f)$ stuff is 'all wet'; here is analysis for $\text{(1)}$ using a fact that the OP might not be aware of, but they can rework their own argument using the definition of continuity at a point.
Proposition: If $f: \Bbb R^n \to \Bbb R$ is a continuous function then
the inverse image of every open interval $(a,b)$ under $f$ is open $\Bbb R^n$.
With $f(x_0) \gt 0$ consider the open interval $I = \big(\frac{f(x_0)}{2}, 2 f(x_0)\big)$. By proposition the set
$\tag 1 U = f^{-1}(I)$
is open and obviously contains $x_0$. So it is a neighborhood of $x_0$. 
Of course when we apply $f$ to any point $x \in U$ we get $f(x) \gt \frac{f(x_0)}{2} \gt 0$.
Since proposition 1 is true when $a,b \in \{-\infty,+\infty\}$, we could also use the interval $I = (0,+\infty)$.
