Are these proofs on partial orders correct? I'm trying to prove whether the following are true or false for partial orders $P_1$ and $P_2$ over the same set $S$.
1) $P_1$ ∪ $P_2$ is reflexive?
True, since $P_1$ and $P_2$ contains all the pairs { ($x,x$) : $x$ in $S$}, we know the union also does. Thus $P_1$ ∪ $P_2$ is reflexive. That set of ($x,x$) elements is called Δ, or the diagonal of $S$.
2) $P_1$ ∪ $P_2$ is transitive?
False, since $x≤y$ and $y≤z$ in $P_1$ ∪ $P_2$. How do we know $x≤y$ is from $P_1 and $y≤z$ isn't from $P_2? There's nothing guaranteeing that we have $x≤z$.
3) $P_1$ ∪ $P_2$ is antisymmetric?
False, for all we know, $x≤y$ in $P_1$ and $y≤x$ in $P_2$. That would be the complete opposite of asymetry - symmetry (for those two elements, at least).

1) Your answer is correct.  One can just as easily say something stronger: if $R_1$ is a reflexive relation on a set $X$ and $R_2$ is any relation on $X$ containing $R_1$, then $R_2$ is also reflexive.  
2) Your conclusion is correct, but you should nail it down by giving a specific counterexample.  Can you find one?  It suffices to take $X = \{1,2,3\}$.
3) To be sure, antisymmetric means that if $x \leq y$ and $y \leq x$, then $x = y$.  Again your conclusion is correct but should be justified.  Here an easy example would be to take $P_1$ to be any linear order $\leq$ on a set with more than one element -- e.g. the standard ordering on $\mathbb{Z}$ or $\mathbb{R}$ -- and $P_2$ to be the dual ordering, i.e., $x \leq_2 y \iff y \leq x$.  Then $P_1 \cup P_2$ is the total relation $x R y$ for all $x,y$.  This is not antisymmetric.  
