Why are the rational numbers not continuous?

You are completely right in that the notion of holes depends on what you can put between your numbers. There is always a bigger set. The integers are contained in the rationals, which are contained in the real numbers, but those themselves can be viewed to have many holes, called non-standard real numbers (e.g. infinitesimals).

In order to give the ideas of “holes” and “being without holes” a meaning, you have to define what the total set of your numbers should be. If you say you want all rationals, that is fine. But there are properties which are only fulfilled by bigger sets of numbers.

For example you can easily construct a right triangle with legs of length $1$. Then the length of the hypotenuse will not be a natural number like $1$ nor even a rational number. It is $\sqrt2$, a hole in the rational numbers. The same happens if you want to measure the circumference of a circle with rational radius.

As for the statement in your book, $\sqrt2$ and $π$ are probably numbers they want on their number line. However what exactly the set of numbers on the number line is, depends on how you define it. There is not a right or wrong way. Just more and less useful ones for doing math with.

In analysis now, there are still more things you would like to describe by numbers. For example you want to find zeros of a function. For the real numbers the following property holds:

Let $f: ℝ→ℝ$ be a continuous function and $a<b$ real numbers such that $f(a) < 0 < f(b)$. Then there is a $x$ between $a$ and $b$ such that $f(x) = 0$.

This is not true for the rational numbers. The reason is, you can always make your interval $(a, b)$ smaller and smaller, keeping $f(a) < 0 < f(b)$, thereby closing in on the zero of $f$. That way you get a sequence of (rational or real) numbers $(a_n)_{n\inℕ}$ which is Cauchy (meaning the distance $\vert a_m-a_n\vert$ gets arbitrary small for $m$ and $n$ big enough). But only in the real numbers there will always be some limit $x$ contained in all the intervals $(a_n, b_n)$. This property is called completeness.

I would prefer to keep two pieces of standard jargon separate, and distinguish being a continuum and being continuous.

Continuity is a property of functions. There's a standard "epsilon-delta" explication of the notion of continuous function. And the function $x^2\colon \mathbb{Q} \to \mathbb{Q}$ (for example) defined over the rationals is just as good a continuous function as its counterpart defined over the reals.

So you can have continuous functions defined over rationals; but the rationals are not a continuum. Why not? The notion of a continuum has its roots in geometry, and -- in the barest headline terms -- the fundamental thought is that certain geometric constructions, even repeatedly indefinitely, will have determinate results. In particular, if we "sum" finite lines by laying them end to end, then so long as the result is bounded, then it is indeed another finite line (how trite that sounds!).

But when we arithmetize geometry, this turns into the thought that, for numbers apt for measuring lines in a true continuum, any increasing sequence of numbers-apt-for-measuring-a-line which is bounded above must converge to a limit which is another number-apt-for-measuring-a-line. Now, a bounded above sequence of rationals need not converge to a rational limit (it might converge to $\sqrt{2}$). By contrast, a bounded above sequence of reals will converge to a real limit. So the reals form the analogue of a geometrical continuum and the rationals don't.

(Sometimes it is said that forming a continuum is a matter of not being gappy. But the notion of gappiness is too blunt an instrument here. After all, the rationals are of course dense, i.e. between any two there is another -- so in one good sense they are not gappy. So we need to distinguish lack-of-gaps meaning denseness, and lack-of-gaps meaning being a continuum.)

You argue that the integers are "not continuous", which means that they are a disconnected set. By the same reasoning the rational numbers are disconnected, see http://www.proofwiki.org/wiki/Rational_Numbers_are_Totally_Disconnected/Proof_1.