Proof by contradiction:
By def of $\sup(A)=x$, $x$ is the upper bound of $A$ which means $x$ is an element of $A$. If $\sup A>\sup B$ then $x$ may not be an element of $B$. Hence contradiction as all the elements of $A$ need to be in $B$, as $A$ is a subset of $B$.
A reminder of definitions: For a subset $S$ of a partially ordered set $P,$ an upper-bound is an element $u\in P$ such that for all $x\in S,$ $x\le u.$ An upper-bound $u$ is called a supremum when for all upper-bounds $v$ of $S,$ $v\ge u,$ and is denoted $\sup(S).$
To prove $\sup(A)\le \sup(B),$ it is enough to show, by definition, that $\sup(B)$ is an upper-bound of $A.$ That is, for all $x\in A,$ $x\le \sup(B).$ This is obvious, since $A$ is a subset of $B,$ so in particular, $x\in B,$ and the inequality is true by definition.