I am aware that solutions to this problem are posted here and here. I was hoping that someone could verify my approach here (which is very similar to the answer posted by @Mittens at the first link).
Let $I_{0} = (0, 1)$ and $K_{0}\subseteq I_{0}$ be a fat Cantor set such that $m(K_{0}) = m(I_{0})/2$. Let \begin{equation*} I_{0}\setminus K_{0} = I_{1} = \coprod_{j = 1}^{\infty}I_{1, j} \end{equation*} (a countable disjoint union of open intervals) and for each $j$ let $K_{1, j}\subseteq I_{1, j}$ be a fat Cantor set such that $m(K_{1, j}) = m(I_{1, j})/2$. Define \begin{equation*} K_{1} = \bigcup_{j = 1}^{\infty}K_{1, j}. \end{equation*} In general, let \begin{equation*} I_{n}\setminus K_{n} = I_{n+1} = \coprod_{j = 1}^{\infty}I_{n+1, j} \end{equation*} and for each $j$ let $K_{n+1, j}\subseteq I_{n+1, j}$ be a fat Cantor set such that $m(K_{n+1, j}) = m(I_{n+1, j})/2$. Define \begin{equation*} K_{n+1} = \bigcup_{j = 1}^{\infty}K_{n+1, j}. \end{equation*} Define $K_{\text{even}} = \bigcup_{n = 0}^{\infty}K_{2n}$ and let $K_{\text{odd}} = \bigcup_{n = 0}^{\infty}K_{2n+1}$ and note that they are disjoint. We claim that $K_{\text{even}}$ is the required set.
Let $I\subseteq I_{0}$ be an open interval (without loss of generality). Since $\sup_{j}m(I_{n, j})\searrow 0$ as $n\rightarrow\infty$ there exists a least integer $m\geq 0$ such that $I\not\subseteq I_{m+1}$. This means that $I\subseteq I_{m, j_{0}}$ for some $j_{0}$ but $I$ is not contained in any component interval of $I_{m, j_{0}}\setminus K_{m, j_{0}}$. Without loss of generality, $m$ is even (since a similar argument holds for the case when $m$ is odd). By the usual construction of the Cantor set, let $K_{m, j_{0}} = \bigcap_{\ell = 0}^{\infty}\coprod_{i = 1}^{2^{\ell}}C_{\ell, i}$ where each $C_{\ell, i}$ is a closed interval. Once again, since $m(C_{\ell, i})\rightarrow 0$ as $\ell\rightarrow\infty$, there exist $\ell_{0}$ and $1\leq i_{0}\leq 2^{\ell_{0}}$ such that an endpoint $x$ of $C_{\ell_{0}, i_{0}}$ is contained in $I$. This shows that there exists a sufficiently large $\ell_{1}\geq \ell_{0}$ and $1\leq i_{1}\leq 2^{\ell_{1}}$ such that $C_{\ell_{1}, i_{1}}\subseteq I$ (with $C_{\ell_{1}, i_{1}}$ having $x$ as an endpoint). Note that there exist indices $j_{1}$ and $j_{2}$ such that $I_{m+2, j_{2}}\subseteq I_{m+1, j_{1}}\subseteq C_{\ell_{1}, i_{1}}$. Therefore,
\begin{align*}
m(I\cap K_{\text{odd}})\geq m(I\cap K_{m+1, j_{1}}) = m(K_{m+1, j_{1}}) = \frac{1}{2}m(I_{m+1, j_{1}}) > 0.
\end{align*}
and
\begin{align*}
m(I\cap K_{\text{even}})\geq m(I\cap K_{m+2, j_{2}}) = m(K_{m+2, j_{2}}) = \frac{1}{2}m(I_{m+2, j_{2}}) > 0.
\end{align*}
Finally,
\begin{equation*}
m(I)\geq m(I\cap K_{\text{even}})+m(I\cap K_{\text{odd}}) > m(I\cap K_{\text{even}}).
\end{equation*}
Any help is appreciated. Thanks a lot!
