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??

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.
