Formal proof that diameter of subset is bounded by diameter of superset (in metric space) I am asked to show that if $A \subset B$, then $\delta(A)\leq \delta(B)$, where $\delta(A)=\sup_{x,y \in A}d(x,y)$ is the diameter for the non-empty set A in the metric space $(X,d)$.  
The fact that it is so intuitive is blocking me from writing a formal proof; OF COURSE if we take the supremum across a subset it's going to be at least as big for the superset...  Is there something I'm missing?
My pathetic proof thus far:
$\delta(A)=\sup_{x,y \in A}d(x,y) \leq \sup_{x,y \in A \cup B }d(x,y) = \sup_{x,y \in B}d(x,y) = \delta(B)$

This follows from $\sup X\le \sup Y$ for $X\subseteq Y$. Which again follows from $\forall x\in Y\colon x\le s\implies \forall x\in X\colon x\le s$
