From Spivak's Calculus:
Consider the function $f(x)=\sin(\frac{1}{x})$. It is easy to see that, so long as $\epsilon < 1$, there will not be one $\delta > 0$ that works for these functions at all points $a$ in the open interval $(0,1)$.
In order to show that there will not be one $\delta > 0$ that works for these functions, I did:
$|f(x)-f(y)|=|\sin(\frac{1}{x}) - \sin(\frac{1}{a})|\le \dfrac{|x-a|}{|x||a|}<\dfrac{2\delta}{a^{2}}$,
since $\biggl|\sin(\frac{1}{x}) - \sin(\frac{1}{a}) \biggr|= 2\biggl|\cos\dfrac{\dfrac{1}{x}+\dfrac{1}{a}}{2}\sin\dfrac{\dfrac{1}{x}-\dfrac{1}{a}}{2}\biggr|=\biggl|2\cos\dfrac{\dfrac{x+a}{xa}}{2}\sin\dfrac{\dfrac{a-x}{xa}}{2}\biggr|$=
$2\biggl|\cos \dfrac{x+a}{2xa}\sin \dfrac{x-a}{2xa}\biggr|\le 2\biggl|\sin\dfrac{x-a}{2xa}\biggr|\le 2\dfrac{|x-a|}{2|xa|}=\dfrac{|x-a|}{|x||a|}$
Then by choosing $\delta < \dfrac{|a|}{2}$ and using the triangle inequality we get that $\dfrac{1}{|x||a|}<\dfrac{2}{a^{2}}$. Then $\delta = a^{2}\epsilon/2$ would mean that $|f(x)-f(a)|<\epsilon.$ If instead we considered points $a$ in the interval $(0,1]$, would there be any significant changes?