The rough idea is that if there are sets of arbitrarily small measure, then you can define unbounded integrable functions by making them get larger and larger but on subsets that are more quickly getting smaller and smaller. This is why $L^1$ is not contained in $L^\infty$ in the case of $(0,1)$ with Lebesgue measure, for example.
I'll try to give you a little more of the idea without giving everything away. You can prove this with contraposition, so your assumption will be that for all $c\gt0$ there exists a Borel set $A$ with $0\lt\mu(A)\lt c$. Use this to construct an infinite sequence of disjoint Borel sets with positive measure converging to zero, say at a rate faster than $2^{-n}$. Then define a function whose value on the $n^\text{th}$ set is $n$.
All the work is in constructing the sequence of sets. I'd rather not spoil your fun by saying more, unless you get stuck along the way and have a specific question.
December 15 update: To enjoy the new spoiler feature, I decided to post a sketch of one way to construct the sequence in the box below.
Suppose that for all $c\gt0$ there exists a Borel set $A$ with $0\lt\mu(A)\lt c$. Start with any Borel set $A_0$ such that $0\lt\mu(A_0)\lt \infty$. (Technically the finiteness is redundant here, because $\mu$ is assumed to be finite.) For each positive integer $n$, the hypothesis guarantees that given $A_{n-1}$ with positive measure, there is a Borel set $A_n$ such that $0\lt\mu(A_n)\lt \frac{1}{3}\mu(A_{n-1})$. Do this for every positive integer (countable AC). Then define a sequence of Borel sets $B_0,B_1,\ldots$ by $B_n=A_n\setminus\cup_{k\gt n}A_k$. The inequality $\mu(A_k)\lt 3^{n-k}\mu(A_n)$ for all $k\gt n$ implies that each $B_n$ has positive measure (at least half of $\mu(A_n)$) and that $\sum_{n\geq 0}n\mu(B_n)$ converges. If $m\gt n$, then $B_m$ is contained in $A_m$, while $B_n$ is contained in $A_n\setminus A_m$, so $B_m$ and $B_n$ are disjoint. The function $\sum_{n\geq1}n\chi_{B_n}$ is thus in $L^1\setminus L^\infty$.