Isolated Schauder Basis for a Normed Space 
If $\left(e_{n}\right)_{n}$ is a Schauder basis for a normed space $(V,\|\cdot\|)$ prove that each $e_{n}$ is isolated.

I have this question in my exercise sheet and I'm not sure how to prove it. In the solution it says we use the continuity of $f_n(x) = x_n$, where $x = \sum_{n=1}^\infty x_n e_n$, but I'm not sure how to prove the continuity for this functional in a general normed space. Once I have that this is continuous, after assuming the basis is normalized we have that for some $\delta$, $f_k^{-1}(B_\delta(e_k)) \subset R_{>0}$, which means it can't contain any other basis vector, since $f_k(e_i) = \delta_{ik}$.
 A: Well the hint is pretty accurate.
Since $\{e_n\}$ is a Schauder basis, each $x\in V$ is uniquely written as $x=\sum_nx_ne_n$ where $x_n\in\mathbb{C}$. We can therefore define $f_n:V\to\mathbb{C}$ by $f_n(x)=x_n$. This is a linear functional and one can see that it is bounded (but this is not trivial!) Bounded is of course equivalent to continuous, so each $f_n$ is continuous.
Note that obviously $f_n(e_k)=\delta_{n,k}$, the Kronecker delta. Now assume that the set $\{e_n:n\geq1\}$ is not isolated. This means that some $e_k$ from this set has the following property: for every $\delta>0$ there exists an integer $n\neq k$ (where $n$ depends on $\delta$) so that $\|e_n-e_k\|<\delta$.
Since $f_k$ is continuous at $e_k$, we take $\varepsilon=1/2$ and find some $\delta>0$ so that, for all $x\in V$ we have the implication $$\|e_k-x\|<\delta\implies|f_k(e_k)-f_k(x)|<1/2.$$
But by assumption we can find $n\neq k$ so that $\|e_k-e_n\|<\delta$. But then $|f_k(e_k)-f_k(e_n)|<1/2$, i.e. $|1-0|<1/2$, a contradiction.
