Good! You are probably finding it difficult to show that function is not continous at 0 because it is continuous at 0!
Given any $\displaystyle \epsilon> 0$ if x> 0, f(x)= x so we can just take $\displaystyle \delta= \epsilon$. That way, "if $\displaystyle |x- 0|= |x|> \delta$, $\displaystyle |f(x)- f(0)|= |x- 0|= |x|< \delta= \epsilon$". While if x< 0, f(x)= 0 so $\displaystyle |f(x)- f(0)|= |0- 0|= 0< \epsilon$ for any x and so for $\displaystyle |x- 0|= |x|< \delta$.
(The function is not differentiable at x= 0 but is continuous there.)