For (2), each set $I_n$ is chosen specifically so that it has no finite subcover. Indeed, by our assumption that $\{G_\alpha\}$ has no finite subcover, upon dividing $I$ into $2^k$ cells $Q_i$, it must be the case that at least one of these $Q_i$, call it $I_1$, has no finite subcover.
Now suppose we divide $I_1$ into $2^k$ cells $R_j$. It must also be the case that at least one of these $R_j$, call it $I_2$, has no finite subcover, otherwise each $R_j$ is covered by finitely many $G_\alpha$, whereby $I_1$ is covered by finitely many $G_\alpha$, contradicting what we have deduced that $I_1$ has no finite subcover.
Proceeding in this manner, we deduce that each $I_n$ has no finite subcover, which gives us (2).
To prove (3), recall the definition of $\delta$ as the diameter of the set $I$:
$$
\delta = \sup_{x,y\in I}|x-y| = \bigg(\sum_{j=1}^k(b_j-a_j)^2\bigg)^{1/2}.
$$
Now, $I_1$ has half the diameter of $I$ since $I_1$ is one of the cells obtained from $I$ by dividing each edge of $I$ in half. Thus $\operatorname{diameter}(I_1) = 2^{-1}\delta$. Likewise, $I_2$ has half the diameter of $I_1$, so that $\operatorname{diameter}(I_2) = 2^{-1}\operatorname{diameter}(I_1) = 2^{-2}\delta$. Proceeding in this manner, if $n\ge 1$, we deduce that $\operatorname{diameter}(I_n) = 2^{-n}\delta$.
By definition of the supremum, if $x,y\in I_n$, then
$$
|x-y|\le\sup_{x,y\in I_n}|x-y| = \operatorname{diameter}(I_n) = 2^{-n}\delta,
$$
which finally gives us (3).
Best Answer
As an example, look at the 3-cell $I=[0,1]\times[10,20]\times[0,10]$. Then we get $c_1=1/2, c_2=15$ and $c_3=5$. So we can create $2^3=8$ new 3-cells, $$\begin{align*} Q_1 &= [0,1/2]\times[10,15]\times[0,5] \\ Q_2 &= [0,1/2]\times[10,15]\times[5,10] \\ \vdots \\ Q_8 &= [1/2,1]\times[15,20]\times[5,10], \end{align*}$$ whose union is the original 3-cell, $I$.