An orthonormal set cannot be a basis in an infinite dimension vector space? I'm reading the Algebra book by Knapp and he mentions in passing that an orthonormal set in an infinite dimension vector space is "never large enough" to be a vector-space basis (i.e. that every vector can be written as a finite sum of vectors from the basis; such bases do exist for infinite dimension vector spaces by virtue of the axiom of choice, but usually one works with orthonormal sets for which infinite sums yield all the elements of the vector space).
So my question is - how can we prove this claim? (That an orthonormal set in an infinite dimension vector space is not a vector space basis).

Edit (by Jonas Meyer): Knapp says in a footnote on page 92 of Basic algebra:

In the infinite-dimensional theory the term "orthonormal basis" is used for an orthonormal set that spans $V$ when limits of finite sums are allowed, in addition to finite sums themselves; when $V$ is infinite-dimensional, an orthonormal basis is never large enough to be a vector-space basis.

Thus, without explicitly using the word, Knapp is referring only to complete infinite-dimensional inner product spaces.
 A: Take any infinite dimensional inner product space $V$ and any orthonormal sequence $(w_n : n \in \mathbb{N})$. Let $W$ be the subspace generated by this sequence. Then $W$ is certainly an infinite dimensional vector space (because it has an infinite independent subset). Also $W$ has an orthonormal basis, because the inner product on $W$ is inherited from $V$ and thus $(w_n)$ is still an orthonormal sequence in $W$.  This means that the theorem you have suggested, "an orthonormal set in an infinite dimension vector space is not a vector space basis", is not true.
What I believe might be true is that no infinite dimensional complete inner product space has a orthonormal basis. This is the question that Andrey Rekalo addressed in another answer.
A: Edit. The answer assumes that one works in a Hilbert space.
Let $\mathcal A$ denote the set of orthonormal vectors. Then for any orthonormal sequence $(e_n)\subset\mathcal A $,  the element 
$$u=\sum_{n=1}^{\infty}\frac{1}{n}e_n$$
cannot be represented as a finite sum of vectors from  $\mathcal A$.
A: As already mentioned in the other answers and comments, this is true if the space is assumed to be complete (see Andrey Rekalo's answer), but not necessarily otherwise (see Carl Mummert's answer).  In the complete case, more can be said.  If the Hilbert space dimension (cardinality of a maximal orthonormal set) is infinite, then the linear dimension (cardinality of a maximal linearly independent set) is at least $\mathfrak{c}=2^{\aleph_0}=|\mathbb{R}|$.  Of course, this doesn't directly tell you anything if the Hilbert space dimension is $\mathfrak{c}$ or greater, but in the case of separable Hilbert spaces like $\ell^2$ (as defined here), it tells you that not only does an orthonormal set not span, but no subset of cardinality less than $\mathfrak{c}$ can span.  One way to see this is to consider the linearly independent set $\{(1,t,t^2,t^3,\ldots):-1\lt t\lt 1\}\subset\ell^2$.  Since $\ell^2$ imbeds into every infinite dimensional Hilbert space as the closed linear span of any countably infinite orthonormal set, this also demonstrates the general fact.  This and other proofs can be found in the solutions to Problem 7 of Halmos's Hilbert Space Problem Book, which I highly recommend. This statement also extends to Banach spaces.
Strictly speaking, what I have said so far hasn't answered your question in the case of Hilbert space dimension greater than or equal to $\mathfrak{c}$.  However, you can get an overkill solution to your question by taking a countably infinite subset $A$ of an orthonormal subset of an arbitrary Hilbert space $H$.   If $M_0$ is the linear span of $A$ and $M=\overline{M_0}$ is the closed linear span of $A$, then $H=M\oplus M^\perp$, while the linear span of your orthonormal set is contained in $M_0\oplus M^\perp$, which is properly contained in $H$ because $M$ has higher dimension than $M_0$.  ($M^\perp$ denotes the set of vectors orthogonal to every element of $M$, and $\oplus$ is being used here to denote internal direct sums.)  All that was really needed here is that the linear dimension of $M$ is not countable, and this also follows from Baire's theorem.
Andrey Rekalo gives a better, nonoverkill answer for the complete case.
