On the Wikipedia page for dyadic cubes, the article claims that if $\Delta^{\alpha}$ denotes all the dyadic cubes in $\mathbb{R}^{n}$ translated by a vector $\alpha\in\mathbb{R}^{n}$, then
There is a universal constant $C>0$ such that for any ball with radius $r<1/3$, there is an $\alpha\in\left\{0,1/3\right\}^{n}$ and a cube $Q$ in $\Delta^{\alpha}$ containing $B$ whose diameter is no more than $Cr$.
This seems false. In one dimension, consider the interval $I=[-\epsilon/2,1/3+\epsilon/2]$ for $1\gg\epsilon>0$. For $\epsilon>0$ sufficiently small, any such interval $J$ must have $\left|J\right|=2^{l}\geq 1/2$. The only dyadic intervals and $1/3$-translate dyadic intervals which intersect and $I$ and satisfy $\left|I\right|\geq 1/2$ are of the form $[0,2^{l}), [-2^{l},0)$ and $[1/3,2^{l}+1/3), [-2^{l}+1/3,1/3)$, none of which contain $I$.
In fact, playing around with the $\epsilon$, it seems that for any length strictly greater than $1/3$, we can find an interval $I$ which is not contained in a cube in $\Delta^{\alpha}$ for $\alpha\in\left\{0,1/3\right\}$.
If $\left|I\right|\leq 1/3$ (i.e. $r\leq 1/6$), then I believe the claim above is true. Let $l\in\mathbb{Z}$ be such that
$$2^{l}\in\begin{cases}(4\left|I\right|,8\left|I\right|], & {\left|I\right|\leq1/8} \\ [1], & {1/8<\left|I\right|\leq 1/3}\end{cases}$$
Suppose that $I$ is not contained in any dyadic interval of generation $l$. Then there exists an endpoint $x=2^{l}p$ of a dyadic interval $J$ which belongs to the interior of $I$. Since $\bigcup_{m\in\mathbb{Z}}[m+1/3,m+1+1/3)=\mathbb{R}$ and $2^{l}\leq 1$, it follows that there is a $J^{*}\in\Delta^{1/3}$ such that $x\in J^{*}$. If $y=q2^{l}+1/3$ is an endpoint of $J^{*}$, then
$$\displaystyle\left|x-y\right|=2^{l}\left|\dfrac{2^{-l}}{3}+q-p\right|\geq\dfrac{2^{l}}{3}\displaystyle\begin{cases}= 1/3, & {1/8<\left|I\right|\leq 1/3} \\ >\left|I\right|, & {\left|I\right|\leq 1/8}\end{cases}$$
since $q-p$ is an integer, $-l\geq 0$, and $2^{-l}/3$ is never an integer. Since $\left|x-z\right|<1/3$, for any endpoint $z$ of $I$, it follows that $I\subset J^{*}$.
I don't think any of this really matters, though, for application of this "trick". For example with maximal operators, it is enough by scaling and monotone convergence to consider supremums over all balls or cubes of diameter $\leq R$, for some choice of $R>0$.
You are quite right. The author must have thought of diameter when writing "radius".
Indeed, $0$ is not an interior point of any dyadic interval, and $1/3$ is not an interior point of any $1/3$-translated dyadic interval. So it's necessary to rule out the intervals containing both $0$ and $1/3$ as interior points.
You are also right in that the precise number does not matter. The way I see it, what one really needs is the existence of $\gamma<1$ such that for any $k\ge 0$, the family of intervals $$\mathcal F_k = \{ 2^{-k}[n,n+1] : n\in\mathbb{Z}\}\cup \{1/3+ 2^{-k}[n,n+1] : n\in\mathbb{Z}\}$$ forms a "$\gamma$-padded cover" of the line, meaning that $\bigcup_{Q\in \mathcal F_k } \gamma Q = \mathbb{R}$.
(For negative $k$, i.e., larger cubes, one naturally needs to translate by larger amounts.)