Suppose that $f: [0, 1) \to \Bbb{R}$ is uniformly continuous. Show that there exists $N \in \Bbb{N}$ such that for every $i \in$ {$1, 2, ..., N$} and every $x, y \in [\frac{i-1}{N}, \frac{i}{N}), |f(x) - f(y)| < 1$.
And thus show that a uniformly continuous function on $[0, 1)$ must be bounded.
By the definition of uniform continuity, applied with $\epsilon=1$, there's some $\delta>0$ such that changing the value of $x$ by less than $\delta$ results in changing the value of $f(x)$ by less than $1$. Fix a positive integer $N>1/\delta$. You can go from 0 to any point in $[0,1)$ in at most $N$ steps, each of length $\leq 1/N<\delta$. As you take these $N$ or fewer steps, $f(x)$ changes by less than $1$ at each step and therefore by $<N$ altogether. Therefore, at any point in $[0,1)$, the value of $f(x)$ is less than $f(0)+N$ and greater than $f(0)-N$. So $f$ is bounded on $[0,1)$.