Suppose $R,S$ are transitive relations over $A$, prove that $R\cap S$ is transitive.
Let $x,y,z\in A$, since $R,S$ are transitive then $$(x,y),(y,z),(x,z)\in R \wedge S\Rightarrow (x,y),(y,z),(x,z)\in R \cap S$$ So $R\cap S$ is transitive.
My question is, why this isn't the right approach?
The OP clarified that $(x,y),(y,z),(x,z)\in R \wedge S$ is an abbreviation of $(x,y),(y,z),(x,z)\in R\land (x,y),(y,z),(x,z)\in S$.
Let $x,y,z\in A$.
While $(x,y),(y,z),(x,z)\in R \wedge S\Rightarrow (x,y),(y,z),(x,z)\in R \cap S$ is actually true, it doesn't prove that $R\cap S$ is transitive.
Recall that, by definition, a relation $T$ on $A$ is transitive if, and only if,
$$\forall x,y,z\in A(((x,y)\in T\land (y,z)\in T)\implies (x,z)\in T).$$
This doesn't look what you've written.
You want to prove that $((x,y)\in R\cap S\land (y,z)\in R\cap S)\implies (x,z)\in R\cap S$. As usual, assume the antecedent and try to get the consequent.