An interesting question, that would take many pages to begin to answer! We make a small disjointed series of comments.
In the last few years, there has been a systematic program, initiated by Friedman and usually called Reverse Mathematics, to discover precisely how much we need to prove various theorems. The rough answer is that for many important things, we need very much less than ZFC. For many things, full ZF is vast overkill. Small fragments of second-order arithmetic, together with very limited versions of AC, are often enough.
About the Axiom of Choice, for a fair bit of basic analysis, it is pleasant to have Countable Choice, or Dependent Choice, at least for some kinds of sets. We really want, for example, sequential continuity in the reals to be equivalent to continuity. One could do this without full DC, but DC sounds not unreasonable to many people who have some discomfort with the full AC. This was amusingly illustrated in the early $20$-th century. A number of mathematicians who had publicly objected to AC turned out to have unwittingly used some form of AC in their published work.
Next, bases. For finite dimensional vector spaces, there is no problem, we do not need any form of AC (though amusingly we do to prove that the Dedekind definition of finiteness is equivalent to the usual definition.)
For some infinite dimensional vector spaces, we cannot prove the existence of a basis in ZF (I guess I have to add the usual caveat "if ZF is consistent"). However, an algebraic basis is not usually what we need in analysis. For example, we often express nice functions as $\sum_0^\infty a_nx^n$. This is an infinite "sum." The same remark can be made about Fourier series. True, we would use an algebraic basis for $\mathbb{R}$ over $\mathbb{Q}$ to show that there are strange solutions to the functional equation $f(x+y)=f(x)+f(y)$. But are these strange solutions of any conceivable use in Physics?
Finally, why should the Banach-Tarski result be unacceptable to a physicist as physicist? It is easy to show that the sets in the decomposition cannot be all measurable. In mathematical models of physical situations, do non-measurable sets of points in $\mathbb{R}^3$ ever appear?
Existence of Schauder bases in separable Hilbert spaces can be proven without any appeal to the axiom of choice, countable or not. Here is the outline.
Let $v_1,v_2,\dots$ be a dense countable subset of the Hilbert space $H$ in question. Let is look at every element $v_i$ of the sequence and look if it is in the linear span of $v_j,j<i$. If it is, we discard it, and if not, we keep it. This way, we get a (possibly finite; for simplicity I'll assume there are infinitely many of them, the proof is the same otherwise) sequence $w_1,w_2,\dots$ of elements of $H$ which form a linearly independent set, and which moreover linearly span a subspace of $H$ containing each $v_i$, so the linear span is in $H$. By applying the Gram–Schmidt process (which is completely constructive, no choice needed) we can conclude there is a sequence $u_1,u_2,\dots$ which is orthonormal and topologically spans $H$. We claim this sequence is a Schauder basis.
Let $v\in H$ be arbitrary. Uniqueness of representation in terms of basis is clear: if $v=\sum_{i=1}^\infty a_iu_i$, then it's easy to see $\langle v,u_i\rangle=a_i$. So we are left with existence. Recall the vectors $v_i$ were dense in $H$, so there is a sequence $v_{i_n}$ convergent to $H$ (no choice needed - e.g. let $i_n$ be the least such that $||v-v_{i_n}||<1/n$. By construction, $v_{i_n}$ is in the linear span of $u_k$, so there is a unique sequence of reals $a_{n,k}$, finitely many of which are nonzero, such that
$$v_{i_n}=\sum_{k=1}^\infty a_{n,k}u_k.$$
The sequence $v_{i_n}$ is Cauchy, hence so is the sequence $a_{n,k}=\langle v_{i_n},a_k\rangle$ for each fixed $k$, thus it converges to some number $b_k$. One can show that then we have $v=\sum_{k=1}^\infty b_ku_k$ (this is very similar to the proof that $\ell_2$ is a Hilbert space), which shows existence. Hence $u_1,u_2,\dots$ is a Schauder basis of $H$.
Best Answer
Yes, you can always require that your objects have the properties you'd need choice for.
Just like you can always require that your sets are well ordered, and simply limit the generality.
The problem is that a lot of times naturally occurring objects will not be well behaved without choice. For example, $\ell_2$ might not have a basis without choice.