Let be $M:=\left\{\omega\in\Omega\mid \liminf\limits_{n\to\infty}X_n(\omega)\geq X(\omega)\right\}$. Show that the complement set $M^c$ satisfies \begin{align*} &M^c=\\ &\bigcup\limits_{m=1}^{\infty}\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}\left\{\omega\in\Omega\mid X(\omega)-X_{j}(\omega)\geq\frac{1}{m}\right\}\\ &\cap \bigcap\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\bigcap\limits_{j=n}^{\infty}\left\{\omega\in\Omega\mid X_{j}(\omega)-X(\omega)\leq\frac{1}{m}\right\}. \end{align*}
My approach:
Let's choose an arbitrary $\omega\in M$ and assume $\liminf\limits_{n\to\infty}X_n(\omega)= X(\omega)$. As $X(\omega)$ is the smallest limit point, for any $\epsilon>0$ there only exist finite many $n\in\mathbb{N}$ such that $X(\omega)-X_n(\omega)>\epsilon$. Otherwise there would be another smaller limit point which is a contradiction. Hence, for any $\epsilon>0$ there must be a $n_0\in\mathbb{N}$ such that for all $n>n_0$ we have $X(\omega)-X_n(\omega)\leq\epsilon$. So in other words \begin{align*} &\omega\in\bigcap\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\bigcap\limits_{j=n}^{\infty}\left\{\omega\in\Omega\mid X(\omega)-X_{j}(\omega)<\frac{1}{m}\right\}. \end{align*}
Now let's choose an arbitrary $\omega\in M$ and assume $\liminf\limits_{n\to\infty}X_n(\omega)-X(\omega)=\delta>0$. As $\liminf\limits_{n\to\infty}X_n(\omega)$ is a limit point, for each $\epsilon$-neighbourhood of $\liminf\limits_{n\to\infty}X_n(\omega)$ there are infinitely many $n\in\mathbb{N}$ such that $X_n(\omega)-\liminf\limits_{n\to\infty}X_n(\omega)<\epsilon$. In particular this holds for any $\epsilon$ with $0<\epsilon<\frac{\delta}{2}$ so that we find infinitely many $n$ such that $X_{n}(\omega)-X(\omega)>\frac{\delta}{2}$. This leads to
\begin{align*} &\omega \in\bigcup\limits_{m=1}^{\infty}\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}\left\{\omega\in\Omega\mid X_{j}(\omega)-X(\omega)>\frac{1}{m}\right\}. \end{align*}
Hence, \begin{align*} &M\subseteq \bigcap\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\bigcap\limits_{j=n}^{\infty}\left\{\omega\in\Omega\mid X(\omega)-X_{j}(\omega)<\frac{1}{m}\right\}\\&\cup \bigcup\limits_{m=1}^{\infty}\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}\left\{\omega\in\Omega\mid X_{j}(\omega)-X(\omega)>\frac{1}{m}\right\}. \end{align*}
The other direction of the subset property seems obvious. Finally, \begin{align*} &M^c=\\ &\left(\bigcap\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\bigcap\limits_{j=n}^{\infty}\left\{\omega\in\Omega\mid X(\omega)-X_{j}(\omega)<\frac{1}{m}\right\}\cup \bigcup\limits_{m=1}^{\infty}\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}\left\{\omega\in\Omega\mid X_{j}(\omega)-X(\omega)>\frac{1}{m}\right\}\right)^c\\ &=\bigcup\limits_{m=1}^{\infty}\bigcap\limits_{n=1}^{\infty}\bigcup\limits_{j=n}^{\infty}\left\{\omega\in\Omega\mid X(\omega)-X_{j}(\omega)\geq\frac{1}{m}\right\}\cap \bigcap\limits_{m=1}^{\infty}\bigcup\limits_{n=1}^{\infty}\bigcap\limits_{j=n}^{\infty}\left\{\omega\in\Omega\mid X_{j}(\omega)-X(\omega)\leq\frac{1}{m}\right\} \end{align*} by law of De Morgan.
Is this correct?