Prove that binary relation $\rho$ on set A is symmetric and anti symmetric iff $\rho \subseteq \Delta_{A}$
Let $\rho \subseteq A \times A$ and $\Delta_{A}$ diagonal.
First guess was proof by contradiction.
$\rightarrow$
Let $\rho$ be symmetric and anti symmetric that means that there is $(a,b) \in \rho \Rightarrow (b,a) \in \rho $ and $(a,b) \in \rho \wedge (b,a) \in \rho \Rightarrow a=b$ assume that $\rho \nsubseteq \Delta_{A}$ that means that there is an element $(a,b) \in \rho$ such that it doesn't satisfy symmetric and anti symmetric.
$\leftarrow$
Assume conversely that $\rho$ is such that $\rho \subseteq \Delta_{A}$ and that $\rho$ is not symmetric and not anti symmetric than there can exist an element $(a,b) \in \rho$ such that $a \neq b$ which means that $\rho \nsubseteq \Delta_{A}$.
Therefore we conclude that $\rho$ is symmetric and anti symmetric iff $\rho \subseteq \Delta_{A}$
Is this correct?
You're making it a bit too hard, you need no contradictions.
Suppose $\rho$ is symmetric and anti-symmetric on $A$. We want to show that $\rho \subseteq \Delta_A$. So let $(x,y) \in \rho$. Then $(y,x) \in \rho$ by symmetry. And then $x = y$ by anti-symmetry. So $(x,y) \in \Delta_A$, and you are done.
If $\rho \subseteq \Delta_A$, let $(x,y) \in \rho$. So $x = y$ (as $(x,y) \in \Delta_A$) and so $(y,x) = (x,y)$ is in $\rho$ as required, so $\rho$ is symmetric. To see that $\rho$ is antisymmetric, we assume $(x,y), (y,x) \in \rho$, but the former alone already gives us $(x,y) \in \Delta_A$, hence $x=y$ as required.