Nonseparable normed space contains an uncountable set of isolated elements Let $(X, ||.||)$ be a normed nonseparable space. Then
$\exists \space \{x_n\}_{n \in I} : |I| > |\Bbb N|$ - uncountable set of isolated (discrete as a subspace) elements of $X$.  
My attempt:
Suppose X does not contain such a set. Then X must at least contain set $R_1$, which consists of non-isolated elements. But then we can extract a dense subset $Q_1 \subset R_1$ (not sure about this part), where $|Q_1| = |\Bbb N|$.
Now let us take a look on $X/R_1$. If it does not contain any subset like $R_1$, then it is countable (it cannot be uncountable set of isolated elements, too), but all countable spaces are separable, hence we must find another $R_2$ with the same condition. But the same way we also find $Q_2 \subset R_2$, $|Q_2| = |\Bbb N|$.
And so on. Now we have families $\{R_n\}_{n \in I}$ and $\{Q_n\}_{n \in I}$. If $|I| = |\Bbb N|$, then X is countable, so $|I| > |\Bbb N|$. Construct a set $\{r_n\}_{n \in I}, \space r_i \in R_i$. This set contains of isolated elements of $X$, and it is uncountable, a contradiction.  
I am afraid that this proof does not work at all, i.e. one should have completely another way to prove this (and be even easier than mine). More than that, I didn't even use a norm of a space. What do you think?
 A: In any metric space $X$, $X$ separable is equivalent to $X$ has no uncountable discrete subspaces (where every point is a relatively isolated point, not an isolated point in $X$ necessarily). I give full proofs here.
A proof could go as follows: suppose that (1) holds:

(1) for every $r>0$ a disjoint family of open (or closed) balls of radius $r$ is at most countable.

Then $X$ is separable: For each $n$, let $\mathcal{D}_n = \{B(x_m, \frac{1}{n}): m \in \mathbb{N}\}$ be a maximal (by inclusion) family of pairwise disjoint balls of radius $\frac{1}{n}$ (Zorn's lemma will show these exist). By assumption (1), this family is indeed countable, hence the indexing by $m \in \mathbb{N}$. Let $D_n = \{x_m: m \in \mathbb{N}\}$ be the set of these centres, and let $D = \cup_n D_n$, which is countable as a countable union of countable sets.
Then we can check that $D$ is dense in $X$, showing $X$ is separable: let $B(x,r)$ be any open ball in $X$, we want to show it intersects $D$. Pick $m$ such that $\frac{2}{m} < r$. We can assume $x \notin D_m$ (or we're done) and so by maximality of $\mathcal{D}_m$ for some member $B(x_m, \frac{1}{m})$ we have $B(x_m, \frac{1}{m} \cap B(x, \frac{1}{m}) \neq \emptyset$. By the triangle inequality this implies $d(x,x_m) < \frac{2}{m} < r$, so that $x_m \in B(x,r)$ and $D \cap B(x,r) \neq \emptyset$, done.
So if $X$ is metric and non-separable, (1) cannot hold: there is some $r>0$ (and thus also for all smaller radii) such that there is an uncountable pairwise disjoint set of $r$-balls. The centres of these balls form an uncountable discrete set, all more than $2r$ apart.
A: We will use the following lemma.

(Riesz's lemma)
  Let $X$ be a normed space, $Y$ be a proper closed subspace of $X$, and $\alpha\in(0,1)$. Then there exists an $x\in X$ with $\|x\|=1$ such that $\|x-y\|\ge \alpha$ for all $y\in Y$.

Proof:
Due to a consequence of the Hahn-Banach theorem, the fact that $Y$ is a proper closed subspace of $X$ implies that there is a functional $f\in X^*$ such that $\|f\|=1$ and $f|_Y=0$.
Let $x\in X$ such that $\|x\|=1$ and $|f(x)|>\alpha$, which is possible because $a<\|f\|=\sup_{\|x\|=1}f(x)$.
Then if $y\in Y$, we have
$$
\|x-y\|=\|f\|\|x-y\|\ge |f(x-y)|=|f(x)|>\alpha. \tag*{$\square$}
$$
Now we prove the claim. Since $X$ is nonempty, there exists $x_1\in X$. Taking $M_1:=\overline{\operatorname{span}\{x_1\}}$, we use the fact that $X$ is nonseparable to deduce that is a proper closed subspace of $X$. Then use the lemma with $\alpha=1/2$ to find $x_2\in X$ such that $\|x_2-x_1\|\ge 1/2$.
Continue inductively...
The above argument can be completed if you are familiar with transfinite induction. (Note that we need to go past the countable case, so normal induction will not suffice.) Alternatively you can use Zorn's lemma.
