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: 
The following perhaps could be a comment, but I thought that maybe I should state it as an answer so it can get positive or negative feedback.

To my mind, whenever I have taught linear algebra in the past, I emphasises that the great power of having a basis B for a vector space V, to study a linear map T from V to some other vector space W it is enough to know what T does on the elements of B. More precisely: T is uniquely determined by its restriction to B; and any function $B \to W$ extends (uniquely) to a linear map $V \to W$.

To make use of this, we therefore want to have examples or results where we can actually specify the values of T on some given basis.

Now suppose you want to study continuous linear maps $V\to W$ between Banach spaces. Since you can never get your hands on a Hamel basis, there is no possibility of creating interesting/useful linear maps by starting with a function defined on that basis. More seriously: if one just works abstractly and says that any function on a Hamel basis extends uniquely to a linear map on the containing space, how would one be able to tell if the resulting linear map is continuous?

Another practical problems is that most of the time, Banach spaces come with extra structure or context: e.g. $L^p(\mu)$ spaces or $C(K)$-spaces. Discontinuous linear maps are just not going to see this structure.

I think that ultimately, the point is that in functional analysis we want to do more than prove results of the form "if such-and-such a map exists, then another kind of map exists". One actually wants to relate Banach spaces to each other: to understand which ones embed as closed subspaces of other ones; to study settings where every bounded linear map from one space to another is automatically (weakly) compact; and so on. Hamel bases will rarely tell us anything here, because they don't reflect or detect various continuity or compactness properties.
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>