Point of discontinuity I have a function: 
$$f(x) = x$$
Defined over the domain $\mathbb{R} \backslash 0$.
Is it correct to say that:
The function is continuous, but it has a point of discontinuity at $x=0$?
 A: No, the continuity or discontinuity of a function at a point is only defined if the point is in the domain.  The function is continuous at every point of its domain, which was stipulated to be $\mathbb R\setminus \{0\}$.  It is not defined at $0$.  
A: The function is continuous for all points where it is defined, which according to you is the set $\mathbb R - \{0\}$. It has no points of discontinuity. A point $x$ is a point of discontinuity for a function $f:D\to \mathbb R$ if the function is defined at that point but its value there is not the same as the limit. When $f$ is not defined at $x$ at all then $x$ can't be considered a point of discontinuity. Think of it this way: the function $f(x)=x^2$, defined on all of the real numbers, is not defined for $x$="the moon". Does it mean that $f$ is discontinuous at the moon?
A: If $x=0$ is not part of the domain of the function, there is no sense in talking about the properties of the function at that point - continuity or anything else.
It happens that the domain on which this function is defined is a part of a larger domain - it is a fundamental issue in mathematics to identify the possibility of extending functions to such larger domains in "good" ways - either preserving useful properties like continuity, or acquiring new ones - like connectedness, compactness or roots to specific equations.
Your $f$ could be extended to $\mathbb R$ by defining $f(0)=\pi$. If you want to preserve continuity, however, you need to define $f(0)=0$.
You might consider the function $g(x)$ defined on the same domain with $g(x)=-x$ when $x$ is negative, and $g(x)=x$ when $x$ is positive. Defining $g(0) = 0$ keeps $g$ continuous, but it is no longer differentiable over the whole domain of definition. Sometimes there is a trade-off between the properties you want and the domain you choose.
A: First, I agree with most other answers: What you say about your $f$ is not totally right but $f$ is continuous where is defined.
I would like to add, that probably some confusion is caused by the fact that there is an obvious extension to a superset of its domain of definition. (Here, set $f(0)=0$, obviously.)
Indeed, the problem if a continuous function defined on some set  has a continuous extension to a larger set is very important in some places of mathematics (e.g. if a continuous linear operator on a Banach space can be extended to a larger Banach space in which the former one is embedded...). Another example is the theorem that a continuous function defined on a dense set of a metric space has a continuous extension to the whole space.
