1.24

First, break the integers into two countably infinite subsets: $$S = \lbrace n \in \mathbb{N} \hspace{0.25cm} | \hspace{0.25cm} n = 2^{m+1} \hspace{0.25cm} \forall m \in \mathbb{N} \rbrace$$ and $$\mathbb{N} \backslash S$$ Next, let $\lbrace a_n \rbrace_{n \in S}$ be an enumeration of $\mathbb{Q} \cap [0,1]^c$, and $\lbrace b_n \rbrace_{n \in \mathbb{N} \backslash S}$ be an enumeration of $\mathbb{Q} \cap [0,1]$.

Certainly, the set $\lbrace r_n \rbrace_{n \in \mathbb{N}} = \big(\lbrace a_n \rbrace_{n \in S}\big) \bigcup \big(\lbrace b_n \rbrace_{n \in \mathbb{N} \backslash S}\big)$ enumerates $\mathbb{Q}$.

Notice the set: $$A = \bigcup_{n=1}^{\infty} \Big( r_n - \frac{1}{n}, r_n + \frac{1}{n} \big) = T \cup V$$ where $$T = \bigcup_{n=1}^{\infty} \Big( a_n - \frac{1}{n}, a_n + \frac{1}{n} \big)$$ and $$V = \bigcup_{n=1}^{\infty} \Big( b_n - \frac{1}{n}, b_n + \frac{1}{n} \big)$$ Observe that $m(A) \leq m(T) + m(V)$, however, by construction, both $m(T) \leq 3$ and $m(V) \leq 2$. (Those numbers, 2 and 3 are both just "safe" over-estimates.) Thus, since $m(A)$ is finite, we have that $A^c$ must be non-empty.