Consider, the function $f:\mathbb Q\to \mathbb R$ defined by
$$f(x)=\begin{cases}1 &\text{ if, } x<\pi\\2 &\text{ if, } x>\pi\end{cases}$$
Show that, $f$ is continuous but NOT uniformly continuous..
Let, $\epsilon >0$ be arbitrary. Then, we can always find a $x>\pi$ and a $y<\pi$ such that $|x-y|<\delta$.
But, $|f(x)-f(y)|=|2-1|=1\not <\epsilon$.
So , $f$ is NOT uniformly continuous..
But I am unable to find that $f$ is continuous....