Why not we avoid the phrase "if we assume AC " and take it as granted?

The question "why worry about the axiom of choice" (or similar) has been asked many times (see here, here, here, or here just to get started). The moral of the story is that there are lots of places where we're interested in doing "set theory" where the axiom of choice fails (even if you aren't a set theorist! For instance many algebraic geometers care about set theory internal to a sheaf topos, which frequently doesn't satisfy AC). Lee Mosher's analogy about groups "assuming the commutative property" is a very good one.

As for the answer to your other questions: Yes, many algebraic properties are true if and only if the axiom of choice is. For instance, it's consistent without AC that

1. There is a ring with a nonzero ideal that is not contained in any maximal ideal (see here for more)
2. There are vector spaces with no basis (see here for more)
3. There are vector spaces over $\mathbb{R}$ with no norm (see here for more)
4. There is a collection of compact spaces whose product is not compact (see here for more, though less than the others)
5. No nonmeasurable subsets of $\mathbb{R}$ exist (see here for more)

I hope this helps ^_^

Given the broad interest in models of set theory where AC fails, it seems more appropriate to clearly state the hypotheses that you are assuming for the model of set theory in which you wish to work, especially for a theorem which is false in certain models where AC fails.

By comparison, if you had a theorem of group theory that was only true "if we assume the commutative property", you would not dare to leave that hypothesis out, given the broad interest in groups that fail to satisfy the commutative hypothesis.

Some of the other answers gave you links to the specific questions you've asked. Both the specific propositions, as well as the historic reasons why we put the axiom of choice on a separate status than, say, the axiom of extensionality or power set.

Let me add a particular answer to the question in your title. Why do we keep stating usage of the axiom of choice, if we take it for granted? Well, easy. We don't take it for granted. Yes, the vast majority of mathematicians today, at least those I came across, tend to take the axiom of choice as a given, but it is not taken for granted.

The reason is the same reason why research into the necessity of the axiom of choice, and fragments thereof, is still quite relevant. The axiom of choice is terrible non-constructive. Even if you accept the law of excluded middle, which in itself is not very constructive, the axiom of choice is still on a different level. It really just tells you things exist, without telling you how these things "look like", in the slightest. How does a Hamel basis of $\Bbb R$ over $\Bbb Q$ look like? Does it contain $\pi$? $e$? $\sqrt 2$? The answers to all of these is yes and no, in the sense that if one exists, then many exist, and there's no reason to prefer one to the other.

So even if you take the axiom of choice as a given, it is still something that prevents you from specifying your objects explicitly. If you want to be able to compute your objects, or present them in a very explicit way, relying on the axiom of choice means you can't do that. Specifying that you're using it hints to the reader that it might not be possible; and studying fragments of the axiom of choice lets us better understand what kind of "oracles" we need in order to do that specific computation.

So, yes. We still mention the axiom because it provides information about how explicit we can get with our understanding of our mathematical objects. But you're right. In some contexts, e.g. topology, we can't really get anywhere without using choice, so we just use it without mentioning it at all. My topology professor told us somewhere around week 7, when we reached compactness, that from this point on we have to assume the axiom of choice, and those who don't want to have nothing more to learn; he was right. (I'm guessing he was not aware of the fragments we did use learning about metric spaces, though. But that's besides the point.)

Finally, let me just remark that $\ell^2$, and indeed any infinite dimensional Banach space, will not have a Hamel basis when working in $\sf ZF+DC+BP$, where $\sf BP$ states "every set of reals has the Baire property", but they are still very useful spaces.