How is it not the case that every continuous function is uniformly continuous?

Question

What's wrong with my following proof?

Suppose $f:\Bbb R\to \Bbb R$ is continuous. Take $x,y\in \Bbb R$.

Given $\epsilon >0$ there exists $\delta_1,\delta_2$ such that $$ 0<|x-x_0|<\delta_1 \rightarrow |f(x)-f(x_0)|<\epsilon\\ 0<|y-x_0|<\delta_2\rightarrow |f(y)-f(x_0)|<\epsilon $$ For some fixed $x_0$.

Then $|f(x)-f(y)|\leq |f(x)-f(x_0)|+|f(y)-f(x_0)|<2\epsilon$ provided $|x-y|<\min\{\delta_1,\delta_2\}$.

So we've found a $\delta$ which works for all $x,y$. This is obviously wrong somewhere, but where?

Answer 1

The $\delta$ you found only works for $x$ and $y$ that are close to $x_0$, so choosing a different $x_0$ means you might need a different $\delta$.

Answer 2

It has to work for every $x_0$! Who would be $x,y$ in the definition of uniform continuity?

Answer 3

Uniform continuity requires you to find, for any given $\varepsilon > 0$, a $\delta$ that works for the entire domain of $f(x)$: The $\delta$-neighborhood of any $x$ maps into the $\varepsilon$-neighborhood of $f(x)$.

That is not what you've shown. You've (more or less) shown that for any given $\varepsilon > 0$ and any given $x$, you can find a $\delta$-neighborhood of $x$ such that any two points in that neighborhood map to points that are within $\varepsilon$ of each other. That isn't really any different from bog-standard continuity.