Here is a proof of the forward implication.
Since $C$ is of content zero, it is bounded and is contained inside a bounded rectangle $Q$. The content of $C$ is the Riemann integral of $f \equiv 1$ over $C$, defined as
$$\int_Cf= \int_Q g,$$
where $g(x) = 1$ for $x \in C$ and $g(x) = 0$ for $x \in Q \setminus C$.
Note that $g$ is Riemann integrable, since it is continuous everywhere except possibly at points of $C$ and the content of $C$ is zero.
Now take any partition $P$ of $Q$. Any subrectangle $R$ of $P$ has non-zero content (by definition of a partition). Hence $R$ is not a subset of $C$ and must contain at least one point where $g(x) = 0$. This implies that $\inf_R g(x) \leqslant 0 \leqslant \sup_R g(x)$. Forming upper and lower Riemann sums we have for any partition $P$,
$$L(P,g) \leqslant 0 \leqslant U(P,g).$$
Since $g$ is integrable we have
$$\int_Q g = \sup_P L(P,g) \leqslant 0 \leqslant \inf_P U(P,g) = \int_Q g,$$
And this implies that $\int_Q g = 0$, i.e., the content of C is zero.
For the reverse implication, if the content of $C$ is zero then $\int_Q g =0$ for any enclosing rectangle $Q$. Hence, for any $\epsilon >0$ there exists a partition $P$ such that $0 \leqslant U(P,g) < \epsilon$. Considering only the subrectangles that intersect $C$, we can show that these subrectangles form a finite covering of $C$ with total volume less than $\epsilon$. See if you can fill in the details of this last step in the proof.