Open cover of $[0,1]$ 
Let $U\cup V=[0,1]$ be an open cover. Then exists $n\in\mathbb{N}$ such that for every $0\leq k<n$ the interval $[\frac{k}{n},\frac{k+1}{n}]$ is completly in $U$ or $V$.

I want to proof this, but I am not sure what exactly is to proof.
[Edit: I gave a proof in the answers, and would be thankful, if one could check it. :)]
At first I understood the question that the intervals $[\frac{k}{n},\frac{k+1}{n}]\subseteq U$ or $[\frac{k}{n},\frac{k+1}{n}]\subseteq V$ where just one statement holds.
But now I think, that it can be $[\frac{k}{n},\frac{k+1}{n}]\subseteq U$ for some $k$ and $[\frac{k}{n},\frac{k+1}{n}]\subseteq V$ for different $k$.
Because that every intervall is just in either $U$ or $V$ seems impossible in general.
For a proof I tried to seperate some cases, but it did not help me to get a clear idea on how to proof this.
Can you give a hint and clear up, what I have to proof?
Thanks in advance.
 A: [Hint]
Certainly you want show that $[\frac{k}{n},\frac{k+1}{n}]\subseteq U$ or $[\frac{k}{n},\frac{k+1}{n}]\subseteq V$ whereby the choice U or V depends on k. The proof will most likely use $[0,1]$ compact, i.e. there are open subsets $U'\subset U$, $V'\subset V$ such that $U',V'$ are unions of finite many intervals and $U'\cup V' = [0,1]$. Maybe you could start proving the statement for these $U',V'$. 
A: This is an alternative way of thinking about this.
Suppose there exists no such $n$. Then for every $n\in\mathbb{N}$ there is some interval $I_n=[\frac{k}{n},\ \frac{k+1}{n}]$ such that $I_n$ is not fully in $U$ or $V$.
In particular, consider the numbers of the form $2^n$ and pick the one such interval, $I_{2^n}$, closest to $0$. This is possible since there are once at a time a finitely many such intervals of interest.
Let $I_{2^n}=[a_n,\ b_n]$. It is easy to see that $\left \{ a_n \right \}$ is increasing. So $a_n \rightarrow a$ for some $a\in [0,\ 1]$.
Now it is that $a\in U$ or $a\in V$, say $U$. Use openness of $U$ and fit some $\varepsilon $-ball into it...
