Let $A=[0,1]\times[0,1], f:A\mapsto \mathbb{R} $ bounded in A. Then the set $D_{f,A}$ of discontinuities of $f$ in $A$ is closed? I think that the statement is true, but I don't know how to start yet. Could anyone help me or give me a hint? I will really appreciate that!

Let
$$f(x,y)=\begin{cases}
\frac 1{q_1q_2}&\mbox{ if }x=\frac{p_1}{q_1}\mbox{ and }y=\frac{p_2}{q_2}\mbox{ are both rational}\\
0&\mbox{ if either is irrational}
\end{cases}$$
Then this is discontinuous on $(\mathbb{Q}\times\mathbb{Q})\cap ([0,1]\times [0,1])$ and continuous on the complement.
