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.
Before going into a formal proof, here is the idea. The space is $\sigma$-finite, so we can "break" it into countably many spaces of finite measure. Up to some technical considerations, we are reduced to the case $X$ of finite measure. An algebra generated by a countable class is countable and we can approximate elements of finite measure by those of a generating algebra.
Let $(A_n,n\geqslant 1)$ be a partition of $X$ into measurable sets of finite measure.
We show that $(A_n,A_n\cap \Sigma,\mu_{\mid A_n\cap \Sigma})$ is separable. Consider $f\in L^p(A_n)$ and fix $\varepsilon>0$. There is $f'=\sum_{j=1}^J a_j\chi_{B_j}$ simple simple such that $\int_{A_n}|f-f'|^p\mathrm d\mu\lt\varepsilon^p$. Define $\mathcal A_n$ the algebra generated by sets of the form $A_n\cap C,C\in\mathcal C$. Then $\mathcal A_n$ is countable. Approximate $B_j$ by $B'_j$, an element of $\mathcal A_n$, that is, such that $\mu(B'_j\Delta B_j)\lt \frac 1{J(|a_j|^p+1)}\varepsilon$. Defining $f'':=\sum_{j=1}^Ja_j\chi_{B'_j}$, we get $\lVert f-f''\rVert^p\lt 2\varepsilon$.
Define $D_n$ as the set of linear combinations with rational coefficients of characteristic functions of elements of $\mathcal A_n$. Since $\mathcal{A}_n$ is countable, so is $D_n$. Finally, define
$$D:=\bigcup_{N\geqslant 1}\left\{\sum_{i=1}^Nd_i,d_i\in D_i\right\}.$$
Then $D$ is countable and dense in $L^p(X)$.
Best Answer
Let $\mu (A) =card (A)$ for all subsets $A$ of $\mathbb R$. Then $L^{p}(\mu)$ consists of all functions $f: \mathbb R \to \mathbb R$ such that $\sum_{x \in \mathbb R} |f(x)|^{p} <\infty$. This space is not separable: supose $(f_n)$ is dense. There exist a countable set $E$ such that $f_n(x)=0$ for all $n$ whenever $x \notin E$. Now let $f(a)=1$ and $f(x)=0$ for $x \neq a$ where $a \in \mathbb R \setminus E$. Then $\|f-f_n\| \geq 1$ for all $n$.