Piecewise constant function continuity In page 291 of "Cracking the GRE Mathematics Subject Test" I came upon the following Topology example:

$$f(x) = \begin{cases} -2, \text{ if } x<0 \\ \;\;\; 2,\text{ if } x \geq 0 \end{cases}$$
Consider the open set $O=(1,3)$, which contains the point $f(0) = 2$. The inverse image of $O$ is the set $[0,\infty)$, which is not open. Therefore the topological definition tells us that $f$ is not continuous at $x=0$.

My question is, can the same argument be used to prove that $f$ is discontinuous for all $x \geq 0$?
From Calculus, I know that $f$ would be discontinuous only at 0. Is it different when using the continuity definition in Topology? Which from Munkres is:

Let $X$ and $Y$ be topological spaces. A function $f:X \to Y$ is said to be continuous if for each open subset $V$ of $Y$, the set $f^{−1}(V)$ is an open subset of $X$.

 A: The given answer seems to imply that continuity at a point is defined like this

A function $f:X\to Y$ is continuous at $x \in X$ if for every open subset $V$ of $Y$ that contains $x$, the set $f^{-1}(V)$ is an open subset of $X$.

and you correctly observe that with this definition, $f$ is not continuous at every $x \geq 0$.
In order to correctly generalize the notion of continuity at a point we need the definition of a neighborhood.

Let $X$ be a topological space and $x \in X$. We say a subset $N$ of $X$ is a neighborhood of $x$ if there is an open set $U$ such that $x \in U$ and $U \subseteq N$.

Then we can define 

A function $f:X\to Y$ is continuous at $x \in X$ if for every neighborhood $N$ of $f(x)$, the set $f^{-1}(N)$ is a neighborhood of $x$.

and you can show that this definition generalizes the metric space definition of continuity at a point, and that a function $f:X\to Y$ is continuous if and only if it is continuous at each $x \in X$.
In the given example, we have that $f^{-1}(O) = [0,\infty)$ is not a neighborhood of $0$, so $f$ is not continuous at $0$, but it is a neighborhood of all $x > 0$.
A: No, it doesn't. For example, consider $f:[a,b] \to \mathbb{R}$ where $f(x) = c$. Naively, you might think that this is problematic, since the preimage of any open set containing $c$ is $[a,b]$ which is a closed subset of $\mathbb{R}$. 
What's the explanation? We're considering $[a,b]$ equipped with the subspace topology, in which it is clopen. 
Therefore, if you restrict your function to the non-negative reals, it is continuous. 
