Exercise Baby Rudin 4.8 Let $f$ be a real uniformly continuous function on the bounded set $E$ in $\mathbb{R}^1$. Prove that $f$ is bounded on $E.$
I am struggling to show that $E$ is compact (I know how to resolve the problem after that). I was originally going to say that it is a subset of a 1-cell since it's bounded but I know that $(0,1)$ isn't compact when $[0,1]$ is. Is it possible to make an argument about limit points?

As pointed out, compactness of $E$ is not needed.
Let $\delta > 0$ be such that $|f(x) - f(y)| < 1$ whenever $x, y \in E$ and $|x - y| < \delta$.
Since $E$ is bounded, you can cover $E$ with finitely many (say, $N$) intervals of length $\delta$. We may also assume that all these intervals actually intersect $E$.
Pick $x_1, \ldots, x_N \in E$ from the each interval. Put $M = \max\{|f(x_1)|, \ldots, |f(x_N)|\} + 2.$
Now, if $y \in E$, then $y$ is in the $\delta$ neighbourhood of some $x_k$. In that case $$|f(y)| \le |f(y) - f(x_k)| + |f(x_k)| < 1 + |f(x_k)| < M.$$
Thus, $M$ is an upper bound.
