Let $\mathbb R$ be a countable union of closed sets, then at least one has nonempty interior Let $\mathbb R=\bigcup_{n=1}^\infty F_n$, I know that a way of proving this is suppose that all the closed sets have empty interior, and recursively construct closed intervals $I_n$ of length less than $\frac{1}{n}$ and $I_1\supset I_2\supset I_3\supset ...$ and $I_n\cap F_n=\emptyset$. But I am not sure how to construct this set and how to finish the proof.
 A: Did you learn and do you remember this proof, due to Cantor, that an interval $[a,b]$ cannot be countable?

Let $r_1$, $r_2$, $r_3$ be a sequence of points that (possibly) may
  include every point of $[a,b]$.  Choose an interval $[a_1,b_1]$ that
  does not include the point $r_1$.  Choose a subinterval of that
  $[a_2,b_2]$ that does not include the point $r_2$.  Just keep doing
  that and use the fact that there must be some point $z\in [a,b]$ that
  belongs to each of the intervals.  But then $z$ is not equal to $r_i$
  for any $i$ so this construction is impossible.

Let's steal this idea from Cantor and apply it here.  [That's how you do mathematical research: steal ideas and twist them around to suit your own purposes.]

Let $F_1$, $F_2$, $F_3$ be a sequence of closed sets with empty
  interiors that (possibly) may include every point of $[a,b]$.  Choose
  an interval $[a_1,b_1]$ that is not included in the set $F_1$.  Choose a
  subinterval of that $[a_2,b_2]$ that is not included in the set $F_2$.
  Just keep doing that ...  construction is impossible.

At each stage you aren't so much "constructing" as claiming the existence of such an interval.  This is exactly your idea and exactly Mr. Ma's hint. Had you remembered Cantor's proof I think you wouldn't be agonizing over the idea quite so much. 
