Are functions still continuous even with removable discontinuities?
Yes, of course. The only thing that matters for continuity are those points at which the function is defined. What occurs outside the domain is irrelevant.
Yes. The function is continuous, because it is still continuous in every point of its (now smaller) domain $(-\infty,2)\cup(2, \infty)$.
I am aware that there are other "definitions" of continuity. The Wikipedia article is confusing in that sense, as it gives a number of definitions that are equivalent to each other, and some that are not:
The most common and restrictive definition is that a function is continuous if it is continuous at all real numbers.
Ugh! Come on, it means that a function not defined on the whole $\mathbb R$ cannot be continuous!
I am sure there are other sources of information that muddle the waters more than they make things clear...
Obviously, if you look at the (continuous) function $f(x)=x, f:\mathbb R\setminus\{2\}\to\mathbb R$ - its graph is not connected, has two components, but that is not what we call discontinuity. A discontinuity would be a point $x_0$ such that $\lim_{x\to x_0}f(x)\ne f(x_0)$, and to even check this condition you must be able to calculate $f(x_0)$, i.e. the function $f$ must be defined in $x_0$ (in our case, $x_0\ne 2$).
Thus, my conclusion for the OP is this:
Take the universally accepted definition of continuity: the function is continuous if it is continuous at every point of its domain. Or any definition equivalent to that one. (I personally like the topological one, where the function is continuous if the preimage of any open subset of the codomain is an open subset of the domain.)
Accept that the graph of a continuous function need not be a connected set (in, say, $\mathbb R^2$) - though, obviously, it will be if the domain of the function itself is connected.
Accept that in maths' education there are other non-equivalent definitions of continuity floating around; fight them with arguments (even if it seems like attacking windmills).