I will focus my answer on the properties which are true for the finite measure spaces but not $\sigma$-finite ones.
Recall Egoroff's theorem:
Let $(X,\mathcal A,\mu)$ a finite measured space, and $\{f_n\}$ a sequence of measurable function from $X$ to $\mathbb R$ endowed with the Borel $\sigma$-algebra. If $f_n\to 0$ almost everywhere then for each $\varepsilon>0$ we can find $A_{\varepsilon}\in\mathcal A$ such that $\mu(X\setminus A_{\varepsilon})\leq\varepsilon$ and $\sup_{x\in A_{\varepsilon}}|f_n(x)|\to 0$.
It's not true anymore if $(X,\mathcal A,\mu)$ is not finite. For example, if $X=\mathbb R$, $\mathcal A=\mathcal B(\mathbb R)$ and $\mu=\lambda$ is the Lebesgue measure, taking $f_n(x)=\begin{cases}1&\mbox{ if }n\leq x\leq n+1,\\\
0&\mbox{otherwise},
\end{cases}$ we can see that $f_n\to 0$ almost everywhere, but if $A$ is such that $\lambda(\mathbb R\setminus A)\leq 1$, then $\mu(A)=+\infty$, hence $A\cap [n,n+1]$ has a positive measure for infinitely many $n$, say $n=n_k$, so $\sup_A|f_{n_k}|\geq \sup_{A\cap [n_k,n_k+1]}|f_{n_k}|=1$.
An explanation could be the following: if $(X,\mathcal A,\mu)$ is $\sigma$-finite, $\{A_n\}$ is a partition of $X$ into finite measure sets, and a sequence converges almost everywhere on $X$, then we have the convergence in measure on each $A_n$: for $k$ and for a fixed $\varepsilon>0$ we can find a $N(\varepsilon,k)\in\mathbb N$ such that $\mu(\{|f_n|\geq \varepsilon\}\cap A_k)\leq \varepsilon$ if $n\geq N(\varepsilon,k)$. The problem, as the counter-example show, is that this $N$ cannot be chosen independently of $k$.
An other result:
Let $(X,\mathcal A,\mu)$ a finite measured space, and $\{f_n\}$ a sequence which converges almost everywhere to $0$. Then $f_n\to 0$ in measure.
We can use the same counter-example as above.
Inclusions between $L^p$ space may change whether the measured space is finite. If $(X,\mathcal A,\mu)$ is a finite measured space, then for $1\leq p\leq q\leq \infty$ we have $L^q(X,\mathcal A,\mu)\subset L^p(X,\mathcal A,\mu)$, as Hölder's inequality shows. But with $X=\mathbb N$, $\mathcal A=2^{\mathbb N}$ and $\mu$ the counting measure, we have for $1\leq p\leq q\leq \infty$, $\ell^p\subset l^q$, so the inclusions are reversed.
Best Answer
$\sigma$-addtivity is not enough. Separability of $L^p(\mu)$ $1\leq p<\infty$ depends and a finer structure of the $\sigma$-algebra $\mathcal{F}$ on which $\mu$ acts, namely countably generated. That is, if there is a countable collection of sets $\mathcal{C}$ such that $\sigma(\mathcal{C})=\mathcal{F}$, then any $L_p$ $1\leq p<\infty$ will be separable. This is the case because simple functions are dense in $L_p$.
Suppose $\mathcal{C}$ is a countable collection of subsets of $\Omega$ and $\mathscr{F}=\sigma(\mathcal{C})$. Let $\{C_n:n\in\mathbb{N}\}$ be an enumeration of the elements of $\mathcal{C}$. For each $n\in\mathbb{N}$ define $\mathcal{A}_n$ the algebra generated by $\{C_1,\ldots, C_n\}$. Then $\mathcal{A}=\bigcup_n\mathcal{A}_n$ is a countable algebra that contains $\mathcal{C}$ and $\sigma(\mathcal{A})=\mathscr{F}$.
For simplicity, assume $\mu(\Omega)<\infty$. Using the procedures of Lebegue-Caratheodory it is easy to check that for any $E\in\mathscr{F}$, there is a sequence $A_n\in\mathcal{A}$ such that $\mu(E\triangle A_n)=\|\mathbb{1}_E-\mathbb{1}_{A_n}\|^p_p\xrightarrow{n\rightarrow\infty}0$ ($1\leq p<\infty$). From this, it is easy to see that $L_p(\mu)$ is separable, for simple functions are dense in $L_p(\mu)$.