$\{Q_i : i\in I\} \text{ is open} \rightarrow U_i:= f^{-1}(O_i)\text{ is open}$

Question

$f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous and $K\subset\mathbb{R}$ is compact.
$\{Q_i : i\in I\}$ is an open cover for $f(K)$.
Does it follow that $U_i:= f^{-1}(O_i)$ is open for each $i \in I$? Is $\{U_i:i\in I \}$ an open cover for $K$?
Intuitively, the answer to both of these questions seems yes. However, it seems conceivable that this is wrong and I would like to know for sure.

Answer 1

Given a continuous function, the preimage of an open set is open.