Condition on Equality of closure of open ball and closed ball, suppose I have a counterexample

Question

Let $(X, d)$ be a metric space. Also for $x \in X$ and $r \ge 0$ define: $$ B(x,r) = \{ y \in X : d(x,y) < r \} \quad \mbox{ and } \quad K(x,r) = \{ y \in X : d(x,y) \le r \}. $$ Denote by $\mbox{cl}(U)$ the closure of some set in the topology induced by $d : X \times X \to \mathbb R$. Then we have

i) $\mbox{cl}(B(x,r)) \subseteq K(x,r)$

ii) if $X$ is normed, then we have $\mbox{cl}(B(x,r)) = K(x,r)$.

This fact is from an authoritative source, guess must people know it? (if not, I can give proof and reference too).

I think I have found a counterexample, so I am a little bit confused.

Counterexample 1: Let $2^{\mathbb N}$ be the set of all one-sided infinite $0$-$1$-sequences. For $\xi = (\xi_i) \in 2^{\mathbb N}$ define $$ || \xi || := \sum_{i=1}^{\infty} \frac{1}{2^i} \xi_i. $$ Then $(2^{\mathbb N}, ||\cdot ||)$ is a normed space over $\mathbb F_2 = \{ 0, 1 \}$ (the finite field with two elements). Therefore it is also a metric space, with metric $$ d(\xi,\eta) = ||\xi - \eta|| = \sum_{i=1}^{\infty} (\xi_i - \eta_i) =\sum_{i=1}^{\infty} \frac{1}{2^i} d'(\xi_i, \eta_i) $$ where $d'(0,0) = d(1,1) = 0, d(0,1) = d(1,0) = 1$ (by the way, this is the discrete metric on $\{0,1\}$), but this inclusion is proper, so both sets are not equal, despite the fact that the space is normed?

Let $\xi = 00000000\ldots$ we have \begin{align*} B(\xi, 1/4) & = \{ \eta : d(\xi, \eta) < 1/4 \} \\ & = \{ \eta = (\eta_i) : \eta_1 = 0, \eta_2 = 0 \} \setminus \{ 001111\ldots \} \end{align*} and \begin{align*} K(\xi, 1/4) & = \{ \eta : d(\xi, \eta) \le 1/4 \} \\ & = \{ \eta = (\eta_i) : \eta_1 = \eta_2 = 0 \} \cup \{ 010000\ldots \} \end{align*} We have $$ \mbox{cl}(B(\xi,1/4)) = \{ \eta = (\eta_i) : \eta_1 = \eta_2 = 0 \} $$ and so $\mbox{cl}(B(\xi,1/4)) \subseteq K(\xi, 1/4)$, but $\mbox{cl}((B(\xi,1/4)) \ne K(\xi, 1/4)$.

What went wrong here?