Here's my attempt at a proof:
Let $A$ be a bounded set in $\mathbb{R}^2$. Then there exists $L>0$ such that $A \subset [-L,L] \times [-L,L]$. Let $\varepsilon > 0$ be given. I intend to cover $A$ by "little" squares of diameter at most $\varepsilon$.
Suppose the edge length of little square be $\varepsilon_{0}$. Since the diameter of squares is nothing but the length of the diagonal, we want $\varepsilon_{0} ^2 + \varepsilon_{0} ^2 < \varepsilon ^2$, i.e, $\varepsilon_{0} < \frac{\varepsilon}{\sqrt{2}}$.
Clearly, $[-L,L]$ is covered by $[-L, -L + \varepsilon_{0}]$, $[-L+\varepsilon_{0} , -L + 2\varepsilon_{0}]$, ... , $[-L+(n-1)\varepsilon_{0}, -L + n\varepsilon_{0}]$ for some $n\in \mathbb{N}$ satisfying $n\varepsilon_{0} \ge 2L$. Now, it is clear that $A \subset [-L,L] \times [-L,L]$ is covered by cartesian product of the covering of $[-L,L]$ and $[-L,L]$. The covers are nothing but the "little" squares that we wanted.
Is this proof correct? Alternative proofs are welcome.
Here is a proof.
Let $\epsilon>0$
Since $A$ is bounded then $\bar{A}$ is compact thus totally bounded,thus exist $x_1,x_2,...,x_m$ such that $\bar{A} \subseteq B(x_1,\epsilon/8) \cup...\cup B(x_m,\epsilon/8)$
Now for $n=1,2,...,m$ take $y_n \in A$ such that $|y_n-x_n| <\frac{\epsilon}{8}$
So $B(x_n,\epsilon/8) \subseteq B(y_n,\epsilon/4)$ and $A \subseteq B(y_1,\epsilon/4) \cup...\cup B(y_m,\epsilon/4)$
In the plane each disk with radius $r$ and center $x$ is the subset of the square centered at $x$ and has diameter $3r$.
So you have the coverings with squares with diameter $\frac{3\epsilon}{4}<\epsilon$