Proving the initial set $0$ is not a separator. This is Exercise 1.16 from "Sets for Mathematics" by Lawvere:
In the category of abstract sets $\mathcal S$, the initial set $0$ is not a separator. (Assume that two sets $A$ and $B$ exist with at least two maps $A\to B$.)
Definitions:

An object $S$ in a category $\mathcal C$ is a separator if and only if whenever $f_1,f_2:X\to Y$ are arrows of $C$ then $(\forall x [S\xrightarrow{x} X \implies f_1 x = f_2 x])\implies f_1=f_2$.


Initial set: There is a set $0$ such that for any set $A$ there is exactly one mapping $0\to A$.

Let $f,g: A\to B$ be maps between sets such that $f\neq g$. Then $f(a)\neq g(a)$ for some $a\in A$. If $h$ is the unique map $0\to A$, then I'm not sure how to proceed from here.
For all $x\in 0$ we have that $fh(x)=gh(x)$, trivially I guess? But I would also suppose that $fh$ can't be equal to $gh$, since $f$ and $g$ disagree on $a$.
 A: The initial object $0$ in the category $\mathbf{Set}$ of sets is the empty set $\emptyset$. For every set $X$, there is a unique map $\emptyset\xrightarrow{!}X$. (This includes $X=\emptyset$, as there is a unique map $\emptyset\xrightarrow{!}\emptyset$.) The symbol $0$ for the initial object in arbitary categories is probably even motivated from this, as $\emptyset$ has zero elements. (For comparison, the terminal object $1$ in $\mathbf{Set}$ is the  set with one element (up to isomorphism, which are bijections in this case).)
For the initial object $S=0$, the condition, that something holds for all morphisms $x\colon S\rightarrow X$ therefore in particular reduces that it holds for the unique morphism $x\colon 0\xrightarrow{!}X$. Taking morphisms $f_1,f_2\colon X\rightarrow Y$, the condition $f_1x=f_2x\colon 0\rightarrow Y$ is always fulfilled, as there is only one unique morphism $0\xrightarrow{!}Y$ due to $0$ being an initial object. Therefore if $f_1\neq f_2$, which can be possible in $\mathbf{Set}$, the conclusion:
$$f_1x=f_2x\Rightarrow f_1=f_2$$
is wrong and therefore $0$ can't be a seperator. (In $\mathbf{Set}$, every nonempty/inhabited set is a seperator though.)
As you see, you would need to have the condition, that you can directly follow $f_1=f_2$ from both morphisms having same domain and codomain. You need a category, where every Hom set only has at most one morphism, a category like this is called thin. Therefore an initial object is a seperator, iff its category is thin. But as you also see, the concept of a seperator is useless in this case as you could directly follow $f_1=f_2$ from both morphisms having same domain and codomain without using a seperator.
