This is an exercise from Munkres', and this is how I did it (I want to know whether my justifications and reasoning were correct): let $E = \{e_n \}_{n \in \mathbb{N}}$, where $e_n$ is defined by $\pi_{i}(e_n) = 1$ if $i = n$ and $\pi_i(e_n) = 0$ if $i \neq n$. I claim that $E $ has no limit points:
First, notice that if $x \in E$, then $x$ can't be a limit point, because $d(x, y) = 1$ for all $y \in E $ with $y \neq x$.
Now, if $x \notin E$, then we can assume that there exists $j \in \mathbb{N}$ such that $d(x, e_j) \in (0,1)$ (otherwise $x$ trivially wouldn't be a limit point). Let $d(x, e_{j_0}) = a$. Then by the triangle inequality, it follows that $d(x, e_i) \geq 1 -a > \min\{\frac{1-a}{2}, \frac{a}{2} \}$ for all $i \in \mathbb{N} $ and it follows that $x$ is not a limit point (note that $x$ being a limit point is equivalent to for all $\epsilon > 0$, there exists $x \neq y \in E$ such that $d(x, y) < \epsilon $ ).
Is that all?
EDIT: Fixed up the proof. It's probably right this time but I'm not sure.
The set $E$ is correct but the justification that it has no limit point in the uniform-metric topology can be better:
Note that $n \neq m$ implies $d(e_n,e_m)=1$ for all $n,m$. In a metric space any open neighbourhood of a limit point of a set $E$ always contains infinitely many points of $E$ (for this we only need that finite sets are closed; metric is overkill but anyway, it's true).
Suppose $p\in [0,1]^\mathbb{N}$ were a limit point, then the open ball $O=B(p, \frac{1}{3})$ would contain infinitely many $e_n$ while $\operatorname{diam}(O) = \frac{2}{3}< 1$ so it cannot even contain two of them...
This contradiction shows that there is no limit point for $E$.