I want to prove that in the category of sets and functions, every epimorphism is a surjective function. My proof: let $f\colon A\to B$ be an epimorphism. Let $u,v\colon B\to C$ be functions. Let $c,c'\in C$, with $c\neq c'$. I define $u(b)=v(b)$ for every $b\in f(A)$, $u(b)=c$ and $v(b)=c'$ for every $b\notin f(A)$. Then $uf=vf$. Since $f$ is an epimorphism, $u=v$, that is false. Contradiction comes from $b\notin f(A)$, so that $f$ must be surjective.
First question: is my proof correct?
Second question: my argument is by contradiction. Is there a direct proof of the same fact?
There are some minor things I would change, such as using $\{0,1\}$ instead of $C$, but it has the general idea.
You have to be a bit clearer about what definition of "surjective" you're using if you want to talk about a "direct proof". For example, if your definition of surjective was "there does not exist an element not in the image", then you've given a direct proof. If your definition is "for all elements in the codomain there is an element in the domain that gets mapped to it", then this is not so direct. You could use the Axiom of Choice to create an inverse which would give a direct means of witnessing the statement. As is well-known, the Axiom of Choice is equivalent to all epimorphisms being split, i.e. having a pre-inverse. That pre-inverse directly provides a witness that there is an element in the domain that gets mapped to any given element of the codomain.