A unit square contains 1 million rectangles without any common points. Show that the total area of rectangles is less than 1. "A unit square contains 1,000,000 rectangles without common points. Show that the total area of rectangles is less than 1."
This statement is somewhat imprecise. Let's say that these are closed rectangles, and "without common points" means that the rectangles are pairwise disjoint. 
This problem seems somewhat trivial to me, but I could be wrong. Is there any formal way to prove this?
 A: Since you have tagged real-analysis, I assume that you can use Lebesgue measure $\lambda$.
Let these rectangles be $(R_j)_{j=0}^{999999}$. Since they are pairwise disjoint, we have
$$ \lambda \left( \bigcup_{j=0}^{999999} R_j \right) = \sum_{j=0}^{999999} \lambda(R_j)
$$
measures also have something called "subadditivity". That is, if $S \subseteq T$, then $\lambda(S) \leq \lambda(T)$. Since all $R_j$s are contained in the unit square $[0,1]^2$,
$$ \lambda \left( \bigcup_{j=0}^{999999} R_j \right) \leq \lambda([0,1]^2) = 1
$$
Hence $\sum_{j=0}^{999999} \lambda(R_j) \leq 1$.
Here is the proof that the above cannot be a equality. For each rectangle $[a_j,b_j] \times [c_j,d_j]$, its complement is open. The set
$$ U = [0,1]^2 \setminus \left( \bigcup_{j=0}^{999999} R_j \right)^c
$$
(the space that is not covered by rectangles) is open. Since open sets have strictly positive measure,
$$ \lambda \left( \bigcup_{j=0}^{999999} R_j \right) = \lambda([0,1]^2 \setminus U) = 1 - \lambda(U) < 1
$$
Edit: Since the unit square is connected, it is not the union of two separated sets. Since all rectangles are closed, they are equal to their closure and so does any of their unions. Thus the rectangles, being separated, cannot cover $[0,1]^2$.
A: Well you can't fill the unit interval with a finite number of pairwise disjoint  line segments (unless the finite number is 1). So you can't fill at least one edge of the unit square with your rectangles. So there's uncovered area.
Edit in response to comment and downvotes - I should have included this picture:

There's no way to cover the full bottom edge with sides of rectangles (not projections). (In fact there will be at least two edges you can't cover since  you can have at most two rectangles each of which covers a full side by itself.)
