Lets begin by noticing that we only have to concern ourselves with $r\in [0, 1]$: any number that can be so approximated will differ by an an integer from a number in that interval. Now, consider the sets
$$A_q = \left[0, \frac{C}{q^3} \right) \cup \left( 1 - \frac{C}{q^3}, 1 \right]\cup \bigcup_{i=1}^{q-1} \left( \frac{i}{q} - \frac{C}{q^3}, \frac{i}{q} + \frac{C}{q^3} \right).$$
Let $x$ be a point satisying the condition, with the sequence $\{q_i\}$ as in the hypothesis. Note that $x \in A_{q_n}$ for all $n$, so we can say that all points satisfying the condition must be in an infinite number of sets of the form $A_q$. Put in terms of sets, this is saying that if$$ B_n = \bigcup_{q=n}^\infty A_q $$then the set of points satisfying the hypothesis in $[0, 1]$ is a subset of $$ X = \bigcap_{n=1}^\infty B_n.$$ To get the result, note that $\lambda(A_q) \leq 2C/q^2$, $\lambda(B_n) \leq 2C \sum_{i=n}^\infty 1/i^2$ and so $$\lambda(X) < \liminf_n B_n = 0$$ as required.