A normed space is infinite dimensional iff there exists an uncountable set without any limit point. A normed space $(X, \|•\|) $ is infinite dimensional iff there exists $S\subset X$ uncountable set without any limit point in $X$.
Proof:
Lemma : If $X$ is a finite dimensional normed space then every uncountable set has a limit point.
Proof is obvious. $S_n=S\cap B[0, n]$
Then $S=\bigcup_{n} S_n$ . At least one of $S_n$ must be infinite, otherwise $S$ would be countable (contradiction!) .
Suppose $S_N$ infinite, then $S_N\subset B[0, N]$.
Since $X$ is finite dimensional, closed unit ball $B[0, 1]$ is compact and scaling map $T_N(x) =Nx$ is a homemorphism on $X$ and thus preserve compact sets.
Then $S_N$ has a limit point, implies $S$ has a limit point.
Hence for a normed space $X$ , if there exists a uncountable subset without any limit point then $X$ must be infinite dimensional.
$•$ Suppose $X$ is infinite dimensional. Then how to construct an uncountable set without any limit points.
Proof or any counter example??
 A: I believe that, in a separable Banach space, every uncountable set $S$ has a limit point: Let $A$ be a countable dense set. Then $S=\bigcup_{a\in A} S\cap B(a,1)$ so that there is $a_1\in A$ such that $S\cap B(a_1,1)$ is uncountable. Recursively, one gets a sequence $a_n\in S\cap \bigcap_{k<n} B(a_k,1/k)$ such that $S\cap B(a_n,1/n)$ is uncountable. The sequence of the centres $a_n$ is then Cauchy and the limit of this sequence is a limit point of $S$.

You probably know that in the non-separable space $\ell^\infty$ of all bounded sequences, the set $S=\{1_A: A\subseteq \mathbb N\}$ (with $1_A(n)=1$ if $n\in A$ and $0$ else) is an uncountable set without limit point.
A: Let $X$ be a  normed linear space and let $Y$ be an infinite subset of $X$ with no limit points. For $y\in Y$ let $\delta (y)=\inf \{\|y-y'\|: y\ne y'\in Y\}.$ We have $\delta (y)>0.$
Claim: If $y_1,y_2\in Y$ with $y_1\ne y_2$ then $B(y_1,\delta (y_1)/3)\cap B(y_2,\delta (y_2)/3) =\emptyset.$
Proof of claim: By contradiction suppose $z\in B(y_1,\delta (y_1)/3)\cap B(y_2,\delta (y_2)/3).$ Then by definition of $\delta (y)$ and by the triangle inequality we have $$\delta (y_1)\le \|y_1-y_2\|\le \|y_1-z\|+\|z-y_2\|<\delta (y_1)/3+\delta (y_2)/3$$  so we have $$(*)\quad \delta (y_1)<\delta (y_1)/3+\delta (y_2)/3 .$$ Interchanging $y_1$ and $y_2$, we also obtain $$(**) \quad \delta (y_2)<\delta (y_2)/3+\delta (y_1)/3.$$ Adding $(*)$ and $(**)$ we have $$0<\delta (y_1)+\delta (y_2)<2\delta (y_1)/3+2\delta (y_2)/3$$ which is absurd.
Now if $Y$ is uncountable then $\{B(y,\delta(y)/3):y\in Y\}$ is an uncountable family of pairwise-disjoint open subsets of $X.$ This is not possible in a separable space. So if $X$ is separable then $Y$ must be countable.
Footnote. There are many infinite cardinals associated with any topological space $S,$ known as topological cardinal functions. The weight $w(S)$ is the least infinite cardinal $k$  such that $S$ has a base (basis) $B$ with $|B|\le k.$ The density $d(S)$ is the least infinite cardinal $k$  such that $S$ has a dense subset $D$ with $|D|\le k.$ The cellularity $c(S)$ is the least infinite cardinal $k$  such that if $O$ is any discrete open family in $X$ (i.e. if $O$ is any pairwise-disjoint family of open subsets of $X$) then $|O|\le k.$ We always have $w(S)\ge d(S)\ge c(S).$ If $S$ is a normed linear space, or any metric space, then $w(S)=d(S)=c(S).$
