Question: I know that Hamel bases have a couple of defects for the purposes of doing analysis in infinite dimensions:

(1) Every Hamel basis of a complete normed space must be uncountable.

(2) For every Hamel basis of a complete normed space, all but finitely many of the coordinate functionals are discontinuous.

I also know both (1) and (2) are false if completeness is dropped.

Here are my questions, which I have labeled A,B,C,D:

A. I don't see why (1) is a serious problem. An uncountable Hamel basis seems just as hard (or just as easy) to work with as a countable Hamel basis. After all, Hamel bases are about unique representation as *finite* linear combinations. What am I missing?

B. (2) seems like it could be inconvenient. But I don't have a concrete understanding of why. What does (2) stop you from doing? I see that it means the standard dual basis for the algebraic dual is not basis a for the continuous dual.

C. Is there something else that goes wrong with Hamel basis in infinite dimensional spaces? Something, perhaps, that is more obviously inconvenient?

D. Is there something that goes wrong with Hamel basis in *incomplete* infinite dimensional spaces?

**Edit:** Some commentators have pointed out that Hamel bases cannot be produced explicitly for the most important spaces. I was aware of that, and should have said so. Is there anything else for C, other than (1),(2), and the lack of explicitness?

Answer: 
Historically, bases in functional analysis arose from reflection on a collection of concrete examples, such as trigonometric functions, orthogonal polynomials, spherical harmonics, eigenfunctions of the Laplacian, etc.

What's in common in these examples is that each appeared as a tool to solve a concrete problem by **diagonalizing an operator**. E. g. you know how the string evolves if released starting with a sine-wave shape, and we would like to know what happens in general. So, we decompose a function into Fourier series, which in general has infinitely many terms. It would be, of course, even better to diagonalize the wave operator into nice functions forming a Hamel basis, but, alas, no such thing exists.

From that point of view, uncountablity is not an issure, since e. g. operators with continuous spectra may be thought of "diagonalizable in an uncountable basis". It's the spectral decomposition of the operators that makes bases (in Hilbert spaces, at least) a particularly important tool.
This is an addition to Yemon Choi's answer.

In fact, a real-valued function $f$ defined on a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$, for instance, extends uniquely to ad *additive* function, not to a linear function (it only needs to be linear over $\mathbb{Q}$). This can be seen readily defining $$F(x)=\sum\_{i=1}^n q\_i f(b\_i),$$ where $b\_i\in H$ for every $i$ and $q\_i$ are $n$ rationals such that $\mathbb{R}\ni x=\sum\_{i=1}^n b\_iq\_i$.

This is a classical tool to provide additive, discontinuous real functions, of course exploiting the axiom of choice.

Finally, a Hamel basis is not only "not write-downable": it can be almost as nasty as you want. For instance, you can prove under AC that there is a Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$ and a function $f:H\to\mathbb{R}$ such that, for every real $x$, $f^{-1}(x)$ has nonempty intersection with every closed, uncountable set without containing any of them (= is a Bernstein set). Or you can prove that there exists a Hamel basis which is not measurable *and* does not have the property of Baire, and much more...

All of this is very nicely covered in Chapter 7 of:

* Ciesielski, K. (1997). *Set theory for the working mathematician* (London Mathematical Society Student Texts No. 39). Cambridge University Press. <https://doi.org/10.1017/CBO9781139173131>