One of the Category Theory axioms I've been provided within some class lecture notes is that If $X,Y,A,B$ are objects of $\mathcal{C}$ and $(X,Y) \neq (A,B)$ then Hom$_{\mathcal{C}}(X,Y) \cap$ Hom$_{\mathcal{C}}(A,B) = \emptyset$.
So I want to verify this axiom for $\mathsf{Set}$, where the morphisms are functions between sets, then for sets $X,Y,A,B$ we let Hom$_{\mathcal{C}}(X,Y) \cap$ Hom$_{\mathcal{C}}(A,B) \neq \emptyset$ and I pick $f$ from this intersection. So $f$ takes values from $X$ and $A$, but I can only say from this either $A \subseteq X$ or $X \subseteq A$, but not and, therefore not giving me equality.
Is there a way to resolve this? Can someone help me out?
$\DeclareMathOperator{Hom}{Hom}$First: saying that $f$ is a function from $X$ to $Y$ means that $f$ assigns an element of $Y$ to every element $x \in X$. Indeed, the domain of a function is entirely determined by the function. So assuming $f \in \Hom_{\mathsf{Set}}(X,Y) \cap \Hom_{\mathsf{Set}}(A,B)$ tells you that $X = A$ on the nose.
Second: in order to prove that these hom-sets are disjoint, you'll need to know exactly what they are. Saying "$\Hom_{\mathsf{Set}}(A,B)$ is the set of functions from $A$ to $B$" is not a precise definition until you've defined what a function is. Once you have a correct, precise definition of function, it'll be fairly straightforward that any function determines its domain and codomain (hence if a function has two domains, they must be the same, and likewise for codomains).
Here's a precise definition you could use in order to define the category $\mathsf{Set}$:
Definition A function from a set $A$ to a set $B$ is a pair $(f,B)$ where $f \subseteq A \times B$ and, for all $a \in A$, there is a unique $b \in B$ such that $(a,b) \in f$.
The ordered pair is really neccesary here, otherwise we would have $\varnothing \in \Hom_{\mathsf{Set}}(\varnothing, \varnothing) \cap \Hom_{\mathsf{Set}}(\varnothing, \{\varnothing\})$ for example.