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?

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 $[0,1]^2$.
