$\bigcup\limits_{i=1}^n A_i$ has finite diameter for each finite $A_i$ Let $(X,d)$ be a metric space. The diameter of a set $A\subset X$ is defined to be
$\operatorname{diam}(A)= \sup\{d(x,y):x,y\in A\}$.
Suppose $A_1, \dots, A_n$ is a finite collection of subsets of $X$ each with finite diameter.
Prove that $\bigcup\limits_{i=1}^n A_i$ has finite diameter.
My approach:
I am trying to show that $\operatorname{diam}(\bigcup\limits_{i=1}^n A_i)\leq \operatorname{diam}(A_1)+\operatorname{diam}(A_2)+\cdots+\operatorname{diam}(A_n)$. If we can show that since each $A_i$ has finite diameter then $\operatorname{diam}(\bigcup\limits_{i=1}^n A_i)$ must also have a finite diameter.
But I stumbled on how to show that inequality, should we use the triangle inequality somewhere? But not very sure how to write it.
Thanks very much!

Hint: Show that $A\subset X$ has finite diameter iff it is bounded. Then, it is easy to see that a finite union of bounded sets is bounded.
