Theorem: The Intersection of 2 anti-symmetric relation is anti-symmetric.
Proof: Let $R_{1}, R_{2}$ be two anti-symmetric relations and $R_{3} = R_{1} \cap R_{2}$
Then $(x,y) \in R_{3} $
$(x,y) \in R_{1} \cap R_{2}$
$(x,y) \in R_{1} \wedge (x,y) \in R_{2}$
$(x \neq y) \rightarrow (y,x) \notin R_{1} \wedge (y,x) \notin R_{2}$
$(x \neq y) \rightarrow (y,x) \notin R_{1} \cap R_{2}$
$(x \neq y) \rightarrow (y,x) \notin R_{3}$
can someone please verify this proof?
It seems like line 4 of your proof should state '$(x,y)\in R_1\land(x,y)\in R_2$' (so that line 5 follows).
Your proof is correct and could be shortened: Note that $(y,x)\notin R_1\implies (y,x)\notin R_3$, so everything concerning $R_2$ isn't needed.
Alternatively: A relation $R$ on a set $A$ is antisymmetric if for all $x,y\in A$, $xRy$ and $yRx$ implies $x=y$.
Suppose $xR_3y$ and $yR_3x$, then $xR_1y$ and $yR_1x$. As $R_1$ is antisymmetric, $x=y$, so we're done.