I am working through basic analysis and tried developing my own proof of this theorem before seeing what the book had to say. I'm sure my proof is wrong but can't see where. Please help me spot what I'm missing - I really want to understand the concepts.
" Suppose $f$ is not uniformly continuous on [a,b]. Then for some $\epsilon > 0$, we can say that no matter what positive $\delta$ we use, take any two points x, y in [a,b] and require the distance between those two points to be less than $\delta$. Even so, the distance between $f(x)$ and $f(y)$ will be more than $\epsilon$. Let $x = a, y=b, \delta = 2|a-b|, \epsilon = 2|f(x) - f(y)|$. But this is a contradiction since $|f(x) - f(y)| < 2|f(x) - f(y)|$. So $f$ is uniformly continuous on [a,b]. "
That is wrong because, even though it is true that, if $f$ is not uniformly continuous, there is some $\varepsilon>0$ such that, for any $\delta>0$, there are $x,y\in[a,b]$ such that $|x-y|<\delta$ and that $\bigl|f(x)-f(y)\bigr|\geqslant\varepsilon$, you have no reason whatsoever to assume that $2\bigl|f(b)-f(a)\bigr|$ is such a $\varepsilon$.