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).
 A: I would not call (2) or (3) a satisfactory answer. 
In order to show something is false, you can not do informal reasoning like "How do we know x≤y is from P1andy≤zisn′tfromP_2? There's nothing guaranteeing that we have x≤z." to show something is false. 
You have to produce two actually partial orderings $P_1$ and $P_2$ such that $P_1 \cup P_2$ is not transitive. 
For example, let $S = \{x, y, z\}$. Let $P_1 = \{(x,y), (x,x), (y,y), (z,z)\}$ and $P_2 = \{(y,z), (x,x), (y,y), (z,z)\}$. Then $(x,y) \in P_1 \cup P_2$ and $(y,z) \in P_1 \cup P_2$ but $(x,z) \notin P_1 \cup P_2$. 
For anti-symmetry, again let $S = \{x,y,z\}$, where all these are three distinct elements. Let $P_1 = \{(x,y), (x,x), (y,y), (z,z)\}$ and $P_2 = \{(y,x), (x,x), (y,y), (z,z)\}$. Then $(x,y) \in P_1 \cup P_1$ and $(y, x) \in P_1 \cup P_2$ however $x$ and $y$ are two different elements. 
A: Good proof of the first one, and your intuition seems to be spot on for the other two, if I'm understanding you correctly. However, you have not proved that a union of partial orders on the same set need not be transitive or antisymmetric.
To do so, you must provide counterexamples. In particular, demonstrate a set $S$ and partial orders $P_1$ and $P_2$ on $S$ such that $R:=P_1\cup P_2$ is not transitive (or antisymmetric, depending on which you're trying to disprove). You should be able to accomplish both with small finite sets.
P.S.: It seems that you might be confusing the definition of antisymmetry. To show that $R$ is not antisymmetric, you must find distinct $x,y\in S$ such that both $x\: R\: y$ and $y\: R\: x.$
A: 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.  
