TITLE: Why isn't every linear subspace of an infinite-dimensional normed space closed?
QUESTION [20 upvotes]: Can somebody please point out the error in the following reasoning? I know that there exist non-closed subspaces of infinite-dimensional normed spaces.
Let $V$ be a vector space of infinite dimension equipped with a norm and $U \subsetneq V$ a proper subspace. $U$ has a basis $\{\hat{u}_\alpha\}_{\alpha \in A}$ and we can extend this basis with additional vectors $\{\hat{w}_\beta\}_{\beta \in B}$ so that $\{\hat{u}_\alpha\} \cup \{\hat{w}_\beta\}$ is a basis of $V$. Let's say that all these basis vectors are normed.
Now I want to prove that $\partial U = U$:
Let $u$ be a vector in $U$ and $\beta \in B$ arbitrary. For any $\varepsilon > 0$ the point $u + \frac{\varepsilon}{2} \hat{w}_\beta$ lies outside of $U$, so $u \in \partial U$.
Let $v \in V \setminus U$, so there exist a $\beta \in B$ and an $a \in \mathbb{R}$ so that $a \,\hat{w}_\beta$ appears in the linear combination that describes $v$. Now for every $\varepsilon < |a|$ the open ball of radius $\varepsilon$ around $v$ doesn't intersect with $U$. As a consequence $v \not\in \partial U$.
Since the boundary of a set is always closed, $U$ has to be closed.

REPLY [20 votes]: Bananach's argument gets to the point. I want to add a bit about a method for identifying the error. Namely, I recommend that you test this kind of arguments against a case where you "know" that a subspace is not closed. Try it with your favorite! Below I walk you thru the first example that came to my mind. You will see that we end up exactly in a kind of situation Bananach described.
Let us consider the space $\ell^2$ of square summable real sequences (with the $L^2$-norm). It has as a subspace the set $V$ of sequences that eventually become constant zero. This subspace is spanned by the sequences
$e_i, i=1,2,\ldots$, defined by declaring that the $i$th term of the sequence $e_i$ is equal to one, but the rest are all zero. You see that $||e_i||=1$.
Let us then pick your favorite sequence from $\ell^2\setminus V$. For example we can use the sequence $x=(x_n)_{n>0}$ where $x_n=2^{-n/2}$. So $x_n^2=2^{-n}$ and therefore by the sum formula for a geometric series $||x||=1$ as well. 
We can just as well include $x$ in the (Hamel) basis of $\ell^2$ in addition to all the sequences $e_1,e_2,\ldots$.
You may already see where we are heading. Fix a (large) integer $k$. The linear combination
$$
x^{(k)}:=x-\frac1{\sqrt2}e_1-\frac1{\sqrt4}e_2-\cdots-\frac1{\sqrt{2^k}}e_k
$$
looks like the sequence $x$ other than that the first $k$ entries are reset to zero. Therefore $||x^{(k)}||^2=2^{-k}$, so when $k\to\infty$ we see that the vectors $x^{(k)}$ become as short as we wish. We also see that $x-x^{(k)}$
is in the subspace $V$. So $x$ is in the closure $\overline{V}$.
But observe that in their respective vector space representations all the vectors $x^{(k)}$ share the coordinate $1$ of the basis element $x$ - even though they become arbitrarily short w.r.t. to our norm. Now reread Bananach's answer!
I might summarize this as:

Infinite dimensional spaces are a bit tricky, and
Hamel bases and coordinates w.r.t. them are not very useful for topological reasoning.

The latter is in sharp contrast to the finite dimensional case (where your intuition understandably comes from at this point of your studies). 
There is a reason why people prefer to write elements of the example space $\ell^2$ as converging series using the "basis" elements $e_i$ as opposed to some Hamel basis (not that I could exhibit you a Hamel basis of $\ell^2$ in the first place - they only exist by virtue of Axiom of Choice).