There exist meager subsets of $\mathbb{R}$ whose complements have Lebesgue measure zero The Baire Category Theorem - Let $X$ be a complete metric space
a.) If $\{U_n\}_1^\infty$ is a sequence of open dense subsets of $X$, then $\bigcap_1^\infty U_n$ is dense in $X$.
b.) $X$ is not a countable union of nowhere dense sets.
The name for this theorem comes from Baire's terminology for sets: If $X$ is a topological space, a set $E\subset X$ if of the first category, or meager, according to Baire, if $E$ is a countable union of nowhere dense sets; otherwise $E$ is of the second category.

Problem 5.3.27 from Folland's Real Analysis: There exist meager subsets of $\mathbb{R}$ whose complements have Lebesgue measure zero

Attempted proof: Let $X$ be a topological space in $\mathbb{R}$ and $E\subset X$ be of the first category (meager). Set $\{x_k\}$ to be an enumeration of the rational numbers, let $$E_n = \bigcup_{k=1}^{\infty} \left(x_k - \frac{1}{2^{k-1} n},x_k + \frac{1}{2^{k-1}n}\right)$$ Consider the set $E = \cap_{1}^{\infty} E_n$, then $$m(E_n) \leq \sum_{k=1}^{\infty}\frac{1}{2^k n} = \frac{1}{n}$$ hence $m(E) = 0$
I am not sure if I am right, any suggestions is greatly appreciated.

Yes, your proof is right. You need to prove that $E^c$ is meager. 
Since $E_n$ is open, $E_n^c$ is closed. $E_n$ is dense in $\Bbb{R}$ for $\Bbb{Q}\subset E_n$. Thus $E_n^c$ is nowhere dense. So $E^c=\bigcup_{n=1}^{\infty}E_n^c$ is meager.
