The argument in your first paragraph is correct.
For the second paragraph, why do you think that being disconnected is of any relevance? The rational numbers are of zero measure because they are countably many of them. The set of irrationals is not countable, therefore it can (and indeed does) have a non-zero measure.
On your third paragraph: It is true that between any two rationals there's an irrational, and between any two irrational there's a rational. However, there is not only one irrational between two rationals, and not only one rational between two irrationals. Rather there are countably many rationals between two irrationals, but uncountably many irrationals between two rationals.
In other words, it's not that you have an alternating sequence of rational and irrational numbers (which is what you seem to have in mind). In particular, you cannot define ”the next larger rational” nor ”the next larger irrational.”
In some sense, although the rationals are already dense, the irrationals are in a sense even more dense.
Maybe it helps if instead of the rationals, you consider the numbers with finite decimal expansion, which also is a countable dense subset. Here, it might be more intuitive that, you have many more possibilities for numbers if you can arbitrarily continue your number string infinitely than if you are forced to stop after finitely many digits. Yet again, between any two finite-expansion numbers, you always can find an infinite-expansion number (just add infinitely many non-zero digits to the smaller one, after first padding it to the length of the larger with zeroes if necessary), and between any two infinite-expansion numbers you can find a finite-expansion number (just cut the larger one at an appropriate point).
It's essentially correct, but it can be simplified.
For every $t=(a_0,a_1,\dots,a_n)\in T_n=\mathbb{Z}^{n+1}\setminus\{(0,\dots,0)\}$, consider
$$
R(t)=\{z\in\mathbb{C}:a_0+a_1z+\dots+a_nz^n=0\}
$$
Then the set $R(t)$ is finite and consists of algebraic numbers. Since every algebraic number belongs to $R(t)$ for some $t\in T_n$ and some $n>0$, the set of algebraic numbers is
$$
\bigcup_{n>0}\,\bigcup_{t\in T_n}R(t)
$$
For every $n$, the set
$$
A_n=\bigcup_{t\in T_n}R(t)
$$
is the countable union of finite sets, so it's countable. Therefore the set $A$ of algebraic numbers is
$$
A=\bigcup_{n>0}A_n
$$
hence a countable union of countable sets.
Note: $A_n$ might in principle be finite (actually it isn't), but this wouldn't invalidate the proof. Read “countable” as “finite or countable”.
Best Answer
Yes, $\mathcal I_A$ is countable. If you want an injection $f:\mathcal I_A\to\Bbb N$, consider a bijection $g:\Bbb N\to A$ and call $$f(x)=\left\lvert\{n\in\Bbb N\,:\, g(n)\in\mathcal I_A\land 0\le n< g^{-1}(x)\}\right\rvert$$
The image of this function is an interval of $(\Bbb N,\le)$ containinig $0$. Its image is a proper subset of $\Bbb N$ if and only if $\mathcal I_A$ is finite (which, apparently, you are not excluding).