[proofplan] We use only the defining order property of the supremum. Since $\sup B$ is an upper bound for $B$ and $A \subset B$, it is also an upper bound for $A$. The least-upper-bound property of $\sup A$ then forces $\sup A$ to be no larger than this upper bound, giving $\sup A \le \sup B$. [/proofplan]

[step:Use inclusion to make $\sup B$ an upper bound for $A$]Since $A$ and $B$ are nonempty and bounded above subsets of $\mathbb{R}$, the least-upper-bound property of the [real numbers](/page/Real%20Numbers) gives the existence of $\sup A$ and $\sup B$. Define \begin{align*} s := \sup A \end{align*} and \begin{align*} t := \sup B. \end{align*} By definition of supremum, $t$ is an upper bound for $B$. Thus, for every $b \in B$, \begin{align*} b \le t. \end{align*} Let $a \in A$. Since $A \subset B$, we have $a \in B$, and therefore \begin{align*} a \le t. \end{align*} Hence $t$ is an upper bound for $A$.[/step]

[guided]Because $B$ is nonempty and bounded above, the real least-upper-bound property ensures that $\sup B$ exists. Likewise, because $A$ is nonempty and bounded above, $\sup A$ exists. We introduce names for these two real numbers: \begin{align*} s := \sup A \end{align*} and \begin{align*} t := \sup B. \end{align*} The goal is to prove $s \le t$. The first point is that $t$ controls every element of $B$, because $t$ is an upper bound for $B$. Thus, for every $b \in B$, \begin{align*} b \le t. \end{align*} Now take an arbitrary element $a \in A$. The hypothesis $A \subset B$ says exactly that every element of $A$ is also an element of $B$. Therefore $a \in B$. Applying the upper-bound property of $t$ to this particular element gives \begin{align*} a \le t. \end{align*} Since the choice of $a \in A$ was arbitrary, every element of $A$ is at most $t$. This is precisely the statement that $t$ is an upper bound for $A$.[/guided]

[step:Apply the least-upper-bound property of $\sup A$] We have shown that $t$ is an upper bound for $A$. Since $s = \sup A$ is the least upper bound of $A$, $s$ is less than or equal to every upper bound of $A$. Applying this to the upper bound $t$ gives \begin{align*} s \le t. \end{align*} Substituting back $s = \sup A$ and $t = \sup B$, we obtain \begin{align*} \sup A \le \sup B. \end{align*} This proves the claim. [/step]