Union of any finite collection of bounded sets is bounded If you could evaluate my proof on scale of 1-5 in clarity, conciseness, and validity, I would appreciate your 2 cent!!! 

Sketch of Proof in words:


  Consider for some $n\in\mathbb{N}$, $\bigcup^{n}_{i=1} P_i$ where $P_i$ is a bounded set in arbitrary metric space $(X,d)$.
   By boundedness, for each $i$, $\exists x^*\in X,\exists M_i\in\mathbb{R}$, $d(x^*,p)\leq M_i$ $\forall p\in P_i$.
  Let $M^*= \max\{M_i:i=1,...,n\}$.
  If $q$ is in the union, then for some $i$, $q\in P_i$.
  $d(q,x^*)\leq M_i\leq M^*$.
   The union of any finite collection of bounded sets is bounded. QED
Assume it is unbounded. It has an unbounded sequence. It has a monotone unbounded subsequence (that is, it goes to infinity). This sequence has infinitely many terms from one of your bounded sets. This set is unbounded.
Assume it is unbounded. It has an unbounded sequence. It has a monotone unbounded subsequence (that is, it goes to infinity). This sequence has infinitely many terms from one of your bounded sets. This set is unbounded.