Prove that the set of all algebraic numbers is countable

Your proof is correct. It skips over a few small details, for example about the fact that you can ask $a_0 \neq 0$ in your $(k + 1)$-tuples, but depending on how nitpicky your grader is I wouldn't worry about that. The only thing I would do is explain, when you say you have a series of nested sets, exactly what those sets are and then be explicit about the theorem you are using to get that their union is countable.

Yes, you're correct; if you want to be more formal, you're using Cantor-Schroeder-Bernstein in your last step http://en.wikipedia.org/wiki/Cantor-Schroder-Bernstein_theorem , in concluding, from the existence of an injection between the set of all roots and the collection of polynomials and an injection between the pairs ($(1,n)$ and the set of all possible roots, that the collection of roots (i.e., the algebraic numbers) is countably-infinite.

You can express the set of algebraic numbers as a countable union of finite sets. You just have to take, for each nonnegative integer n, the set of roots of the polynomials for which k + |a_0| + ... + |a_k| = n, where k is the degree of the polynomial. For every n you only have finitely such polynomials and each polynomial has only finitely many roots.