I'm having trouble with the following proof:
Suppose $\mathcal{R}$ is a partial order on A, $B_1 \subseteq A$, $B_2 \subseteq A$, $\sup(B_1)=x_1$ and $\sup(B_2)=x_2$. Prove that if $B_1 \subseteq B_2$, then $x_1 \mathcal{R}x_2$.
This result seems obvious, but everything I have tried thus far has lead to dead ends. One idea I had is to define $U_1$ as the set of all upper bounds of $B_1$ and and $U_2$ as the set of all upper bounds of $B_2$ and try to show that $U_2 \subseteq U_1$ (this seems to be the case, but I can't find a way to prove this either). Then I could conclude that $x_2 \in U_1$ use the fact that since $\sup(B_1) = x_1$, $\inf(U_1) = x_1$, to conclude $x_1 \mathcal{R} x_2$.
I'm thinking there is a simpler way to go about proving this, but I can't find it. Any help would be appreciated!
Your approach works fine.
If $x\in U_2$, then by definition $b\mathcal{R}x$ for all $b\in B_2$. Let $b\in B_1$ be arbitrary; then $b\in B_2$, so $b\mathcal{R}x$, and since $b\in B_1$ was arbitrary, $x\in U_1$. Thus, $U_2\subseteq U_1$, and you have the rest of the argument.
You can, however, argue a little more directly that if $b\in B_1$, then $b\in B_2$, so $b\mathcal{R}x_2$. Thus, $x_2$ is an upper bound for $B_1$, and hence $x_1\mathcal{R}x_2$. It’s essentially the same argument, of course, but getting more directly to the point.