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.
Best Answer
Your mistake is here: "Now for every $\epsilon <|a|$..." Just because a vector has small norm (here $\epsilon $) its coefficients in a basis do not need to be small. In other words, the linear map sending a vector to one or more coefficients is not necessarily continuous.
You get some intuition for this by looking at two vectors in $ R^2$ that are almost identical, yet independent. Then their difference has a small norm yet relatively large coefficients w.r.t. the basis of the two vectors. In infinite dimensions it can happen that successive basis vectors get more and more "almost dependent", thus giving rise to vectors that have larger and larger coefficients while having a norm bounded by a constant $\epsilon $.
To elaborate more on the relation of your question to continuity of linear maps, consider the map $L\colon V\to W$, $v=(u,w)\mapsto Lv:=w$, where $V=U\oplus W$ is a direct sum decomposition of $V$. If $L$ were continuous, then $U=\ker L$ would indeed be closed. Conversely, when $U$ is not closed, then $L$ is not continuous, which supplements the first paragraph.