every $k$ cell is compact I did not understand the proof from the word "and some sets in this infinite sequence are from $\{G\}$....." to the end. Thanks for helping me to understand.

$k$ cell in $\mathbb{R}^k$ is defined as $\{x=(x_1,\dots, x_k):a_i\le x_i\le b_i\}=[a_1,b_1]\times\dots[a_k,b_k]$
It's not clear what the reasoning is. In my opinion the idea is that you have a sequence of subcells each covered by an infinite cover with no infinite subcover which is itself a subcover of $G$. This sequence converges to a point and this point is contained in an open set $U$ of $G$. But then $U$ must entirely contain a subcell and here it's the contradiction
In that prove we have infinite subcell all of it contain only one point and each of it can be covered by one open set of cover but still $G$ covered by infinite open cover, I think we must say if we can't cover $G$ by finite sub cover we split $G$ into $2^k$ of $k$ cells each of them covered by finite sub cover (one) if one of it can't cover take the same manner and split it (split up finite turn).