My thoughts on proving this statement is as follows:
Suppose $G_a$ is an open cover of $Q= [0,1] \times [0,1]$. For each $x$ in $[0,1]$, there is some ball around $x$ with radius $r_x$ such that it covers $[x-r_x, x+r_x] \times [0,1]$. Since $[0,1]$ is closed and bounded, it is compact. Since $[0,1]$ is compact, this vertical strip described above is also compact, i.e., there is a finite collection of these balls of $G_a$ that covers this vertical strip.
Next we do the same thing except for horizontal strips.
This is where I am drawing a blank, how am I to show that the union of these balls are finite, and how would I know that it covers all of the unit square?
Best Answer
You're on the right track to prove this. For each $x$, let $G_x$ be a finite subcover of $G_a$ that covers $\{x\}\times[0,1]$.
Claim: There exists a positive number $r_x$ for which $G_x$ covers the entire strip $[x-r_x,x+r_x]\times[0,1]$. You need to prove this. One way uses the fact - which also requires proof - that you can assume $G$ consists of rectangles.
With $r_x$ defined in this way for each $x\in[0,1]$, consider the collection of intervals $[x-r_x,x+r_x]$. These cover $[0,1]$, which is compact, so there's a finite set $X$ for which $[0,1]\subseteq \bigcup_{x\in X}[x-r_x,x+r_x]$. Then $\bigcup_{x\in X}G_x$ (which is a finite subcover of $G_a$) covers $Q$.