Is every subset of a partial order that is itself a partial order, a restriction of the original order?

Question

Suppose $P$ is a partial order, that is, a set of ordered pairs that satisfy reflexivity over the field of $P$, anti-symmetry, and transitivity. Suppose $Q$ is a subset of $P$ that is itself a partial order. Then, is $Q$ a restriction of $P$? Meaning, is there some subset $S$ of the field of $P$ such that $Q = P \cap (S \times S)$?

Answer 1

After some thought, I figured out the answer. The answer is no. A counterexample is when $P=\{(1,1),(1,2),(2,2)\}$ and $Q=\{(1,1),(2,2)\}$.