real analysis - A function continuous and bounded on a closed and bounded set but not uniformly continuous there

I'm reading 'Counterexamples in analysis by Bernard Gelbaum' which has this as one the counterexample
Function is defined on $\Bbb{Q}$$\cap$$[0,2]$ which is closed and bounded set

$ f(x) =

\begin{cases}
0, & \text{if 0 $\le$ $x$ $\lt$ $\sqrt2$ } \\
1, & \text{if $\sqrt2$ $\lt$ $x$ $\le$2 }
\end{cases}$

I concluded that it is a continuous function since co-domain is {${0,1}$} and $f^{-1}$ {$0$} and $f^{-1}${$1$} are closed in $\Bbb{Q}$$\cap$$[0,2]$ ( i. e. inverse image of closed sets is closed and hence $f$ is continuous )
but I'm facing trouble in proving that it is not uniformly continuous. Please help.

Also author mentioned that field $\Bbb{Q}$ is not complete and hence we can have such example . If possible then please explain this point also.

Answer

Well some minute points regarding continuity of $f$:

You also need to verify that inverse image of all the opens sets viz.$\{\{0\},\{1\},\{0,1\},\emptyset\}$ are all open .

Regarding uniform continuity of $f$:

Since $\Bbb Q$ is dense ,there exists a sequence $x_n\in [0,\sqrt 2]\cap \Bbb Q$ such that $x_n\to \sqrt 2\implies |x_n-\sqrt 2|<\frac{1}{n}\forall n$.

Similarly since $\Bbb Q$ is dense ,there exists a sequence $y_n\in [\sqrt 2,2]\cap \Bbb Q$ such that $y_n\to \sqrt 2\implies |\sqrt 2-y_n|<\frac{1}{n}\forall n$.

Hence $|x_n-y_n|\le |x_n-\sqrt 2|+|\sqrt 2-y_n|\le \frac{1}{n}\forall n$ but $|f(x_n)-f(y_n)|=1$.

NOTE:Since $\Bbb Q$ is not complete hence it has gaps and always such an example is available