[proofplan] We use the defining bound for the larger set $B$ and restrict it to the subset $A$. Since every point of $A$ is also a point of $B$, the same center and radius that bound $B$ also bound $A$. This proves boundedness without changing the constants. [/proofplan] [step:Restrict the bounding ball for $B$ to the subset $A$] Since $B$ is bounded, there exist a point $x_0\in X$ and a number $R>0$ such that \begin{align*} d(x_0,b)\le R \qquad \text{for every } b\in B. \end{align*} Let $a\in A$ be arbitrary. Because $A\subset B$, we have $a\in B$. Applying the displayed bound with $b=a$ gives \begin{align*} d(x_0,a)\le R. \end{align*} Since $a\in A$ was arbitrary, the point $x_0\in X$ and the radius $R>0$ satisfy \begin{align*} d(x_0,a)\le R \qquad \text{for every } a\in A. \end{align*} Therefore $A$ is bounded. [guided] The only information needed from the hypothesis that $B$ is bounded is the existence of one ball that contains all of $B$. By definition, there are a point $x_0\in X$ and a number $R>0$ such that \begin{align*} d(x_0,b)\le R \qquad \text{for every } b\in B. \end{align*} We now check that this same pair $(x_0,R)$ also works for $A$. Let $a\in A$ be arbitrary. The inclusion $A\subset B$ means precisely that every element of $A$ is an element of $B$, so $a\in B$. Hence the bound valid for all points of $B$ applies to this particular point $a$, and we get \begin{align*} d(x_0,a)\le R. \end{align*} Because no special property of $a$ was used other than $a\in A$, this inequality holds for every $a\in A$. Thus there exist $x_0\in X$ and $R>0$ such that \begin{align*} d(x_0,a)\le R \qquad \text{for every } a\in A. \end{align*} This is exactly the definition of boundedness for $A$, so $A$ is bounded. [/guided] [/step]