Given $\mu$ is semifinite and $\mu(E)=\infty$, prove that if $C>0$, then $\exists$ measurable set $A \subseteq E$ such that $C < \mu(A) < \infty.$

Let ($X,\Sigma, \mu)$ be a measure space. Suppose that $\mu$ is semifinite, $E \in \Sigma$, and $\mu(E)=\infty$. Prove that if $C>0$, then there exists some measurable set $A \subseteq E$ that satisfies $C < \mu(A) < \infty.$

I have a hint:
Define $C =$sup{ $\mu(F)$ $:$ $F \in \Sigma, F\subseteq E, \mu(F) < \infty$}. If $C<\infty$, then there exists measurable sets $F_k \subseteq E$ with finite measure such that $\mu(F_k)\to C$.
A quick definition:
If every set $E \in \Sigma$ with $\mu(E)= \infty$ contains 
an $F \subseteq E$ such that $F \in \Sigma$ and $\mu(F) < \infty$, then $\mu$ is semifinite.
I know the first step is to find a contradiction if C is finite in the hint.

I'm confused as to what the hint is saying: as I have understood, you can't just define $C$ but rather must prove that given $C$, we have $A \subset E$ with $C < \mu(A) < \infty$, no?

https://math.stackexchange.com/questions/234930/if-a-measure-is-semifinite-then-there-are-sets-of-arbitrarily-large-but-finite

Possible duplicate of If a measure is semifinite, then there are sets of arbitrarily large but finite measure

Thanks for the links. Yeah the hint is what confused me.
