Let $K\subseteq \mathbb{R}$ be a non-empty compact set. Let $f:K\rightarrow \mathbb{R}$ be a continuous function, so it is uniformly continuous on $K$ and bounded....Can we show that $K$ can be covered by a finite number of open subsets on which $\lceil f \rceil$ is contant?
So far...here is what I've thought of but I don't know where to go from here:
Since $f$ is continuous then each of the sets $\left\{f^{-1}(n,n+1)\right\}_{n \in \mathbb{Z}}$ form an open cover of $K$. Since $K$ is compact, then there must a finite number of $n_1,\dots,n_k$ such that $\left\{f^{-1}(n_i,n_i+1)\right\}_{i=1}^k$ covers $K$.
No, you cannot show it, since it is not true. Take $K=[0,1]$ (endowed with the standard distance) and $f(x)=x$. Then $\lceil f\rceil(x)=0$ if and only if $x=0$. And $\{0\}$ is not an open set.