Prove that a language is bounded if and only if it's finite
The claim holds only for languages over finite alphabets.

Bounded $L$ $\implies$ Finite $L$

Let $\Sigma$ be the alphabet of $L$ and $L$ be bounded by some $n \in \mathbb{N}$.

The largest possible such $L$, call it $L^\#$, is $\bigcup_{\,i=0}^{\,n} \Sigma^i$ elements. $L^\#$ is finite since $|L^\#| = \sum_{i=0}^{n} |\Sigma|^i$. Therefore, any $L \subseteq L^\#$ must also be finite.

Finite $L$ $\implies$ Bounded $L$

Let $x^\#$ denote the longest string in $L$. Such a string must always exist since $L$ is finite.

Then, $\forall x \in L \ldotp |x| \leq |x^\#|$ and thus, $L$ is bounded.



A simple counterexample to the infinite alphabet case:

Consider an infinite alphabet $\Sigma = \{ s_0, s_1, ... \}$. The language $L = \Sigma$ is bounded since $\forall x \in L \ldotp |x| \leq 1$, but is infinite.