For $(x,y)\in \mathbb{R^{2}}$ with $(x,y)\neq (0,0)$. Let $\theta=\theta (x,y)$ be unique real number such that $-\pi<\theta\leq\pi$ and $(x,y)=(r \cos\theta , r \sin\theta)$ with $r=\sqrt{x^{2}+y^{2}}$, then function $\theta :\mathbb{R^{2}}\setminus\{(0,0)\}\to\mathbb{R}$ is
Clearly, function $\theta$ is defined as $\theta (x,y)=\tan^{-1}\frac{y}{x}$ which is bounded. I am not able to prove that function is not continuous.
This function, called the principal value of the argument, is not "clearly defined" by $\theta(x,y):=\arctan{y\over x}$. In fact this formula is even wrong when $x<0$. A correct definition would be $$\theta(x,y):=\left\{\eqalign{ &-{\pi\over2}-\arctan{x\over y}\qquad(y<0)\cr &\arctan{y\over x}\qquad\qquad(x>0)\cr &{\pi\over2}-\arctan{x\over y}\qquad(y>0)\ ,\cr}\right.\tag{1}$$ and you may add $\theta(x,0):=\pi$ when $x<0$. In any case the function $\theta$ defined in this way is bounded. Furthermore each point $(x_0,y_0)\ne(0,0)$ which is not on the negative $x$-axis has a neighborhood that is completely contained in one of the halfplanes listed in $(1)$. It follows that the function $\theta$ is continuous in all such points. But it is obvious that $\theta$ is not continuous in the points $(x_0,0)$ with $x_0<0$: For such points one has $$\lim_{y\to0-}\theta(x_0,y)=-\pi\ne\pi=\theta(x_0,0)\ .$$
