The function $I_A - I_B$ is equal to $0$ on $A\cap B$ and on $(A\cup B)^c$, and it is equal to $\pm 1$ on $(A\setminus B)\cup (B\setminus A) = A\Delta B$. Thus, $\|I_A - I_B\|_1 = \int_{A\Delta B}1 d\mu = \mu(A\Delta B)$.

Now assume that $A_n\in \mathcal{F}$ for all $n\in\N$ and that $(A_n)$ is a Cauchy sequence in $\mathcal{F}$ with the metric inherited from $L_1(\mu)$, i.e. $f_n := I_{A_n}$ is a Cauchy sequence in $L_1(\mu)$. Since $L_1(\mu)$ is complete, there exists a function $f\in L_1(\mu)$ such that $\|f_n - f\|_1\to 0$. We have to show that $f = I_A$ a.e. for some $A\in \mathcal{F}$.

Since $f_n\to f$ in $L_1(\mu)$, there exists a subsequence $(f_{n_k})_k$ such that $f_{n_k}\to f$ pointwise on a measurable subset $E\subset \Omega$ such that $\mu(\Omega\setminus E) = 0$. But the functions $f_{n_k} = I_{A_{n_{k}}}$ are $\{ 0, 1 \}$-valued, so we conclude that $f(\omega) \in \{ 0,1 \}$ for all $\omega\in E$. Now put $A = [f = 1]$, then $A\in \mathcal{F}$ and we have $f|E = I_A|E$, so $f = I_A$ almost everywhere.