Axiom of choice : continuous function and uniformly continuous How I proof that every continuous function f in [0,1] is uniformly continuous, without axiom o choice? 
I took this from the book Axiom of Choice from Horst Herrilich 

He had a observation that indicates a a proof without axiom of choice but i still dont get it how
 A: The proof given uses choice to select a $\delta_x$ for each $x\in[0,1]$. However, this can be avoided, for the combination of two reasons:


*

*For every $x$ it is possible to select a rational $\delta_x$ with the desired property.

*Even without the axiom of choice there exists a choice function for (nonemepty) subsets of $\mathbb Q$: Since $\mathbb Q$ is known to be countable, take your favorite injection $\mathbb Q\to\mathbb N$ and map each subset of $\mathbb Q$ to the element that has the smallest image under your injection.
Because of (2) you don't need to choose each $\delta_x$ arbitrarily: you can define a deterministic way to choose each $\delta_x$ given $x$, $\varepsilon$ and $f$. If you extend the proof to specify that this deterministic selection of $\delta_x$ is to be used, there is no need for the axiom of choice anymore.
A: The point is that shrinking $\delta_x$ cannot hurt. So we can shrink it to some $\frac1n$ for an appropriately large $n$ without it damaging the proof in any way.
