Can the open unit disk be expressed as the union of a countable collection of closed squares? My question is simple, but I believe it is challenging think of an example. 
Forget the fact that I couldn't think about an example, I couldn't understand the solution to this.
Need some explanation on it.
Answer :
The closed this $D_n$ of radius $1-\frac{1}{n}$ can be covered by a finite number of squares contained in $D$ and union of all $D_n$ would give us $D$
Q1) How come union of all $D_n$ give us $D$
Q2) Aren't we trying to express the unit circle, instead of covering it? 
 A: The open unit disk is  $B_E(0,1)$, a subset of $\mathbb R^2$, where $E$ is the Euclidean metric. We are looking to prove or disprove, that $B_E(0,1)$ can be described with countable unions of $\bar B_M(x_i,r)$, where $M$ is the Manhattan metric, and that the $\bar B$ denotes closed ball. Immediately notice that if the unit disk was closed, there would exist an uncountable union of closed squares: let $T_\theta$ be a transformation that rotates the cordinate system (for example a rotation matrix), and then $M_\theta = M(T_\theta(x),T_\theta(y))$ is a well defined metric. It follows that $$\bar B_E(0,1) = \bigcup_{\theta \in \Theta} \bar B_{M_\theta}(0,1)$$ The following is also true for open unit disk and open squares.
Now for the proof itself: say there does exist $B_E(0,1) = \bigcup_{n \in \mathbb N} \bar S_n$, where $\bar S_n$ is an arbitrary, closed square. Now, $B$ is open and is a union of sets. Therefore the members in the union need to be open - with respect to $E$ - as well. $\bar S_n$ is homeomorphic to $[a,b] \times [c,d]$. Set $U$ is open if for any $x \in U$, there exists $\epsilon >0$ such that, any point $y \in \mathbb R^2$ satisfying $E(x,y)<\epsilon$ means that $y \in U$. Equivalently for every point in $U$, $U$ contains that point's neighbourhood. Take any corner point of $\bar S_n$, in this case $a$: $E(a,y) < 2$ implies we can take $y = a-1$, but $y$ is not in $\bar S_n$, so $\bar S_n$ is not open, and there does not exist a countable collection of closed squares that express the open unit disk. Even more strongly, there does not exists an uncountable collection, as our proof did not rely on the index family $n \in \mathbb N$.
A: Let $S$ be the set of closed squares of non-zero width,   that are subsets of $D.$ Every $p\in D$ lies at the center of (many) members of $S.$ So the set $T=\{$Int$(s):s\in \}$  of interiors of members of $S,$ is an open cover of $D.$
Now $D$ is a separable metric space so it is a Lindelof space. So there exists a countable $T^*\subset T$ such that $\cup T^*=D.$ Then $S^*=\{\bar t: t\in T^*\}$ is a countable subset of $S,$ and $\cup S^*=D.$
