Continuity of a function in two variables
Solution 1:
For $\varepsilon>0$ denote $$X_{\varepsilon} = \Big \{ (x,y) ~|~ \exists \eta >0, \forall (h,k) \in [-\eta,\eta]^2,~ |f(x+h,y+k)-f(x,y)|< \varepsilon \Big\}.$$ Then $f$ is continuous on $\bigcap_{\varepsilon \in \mathbb{Q}_{>0}} X_{\varepsilon}$.
Lemma: For any $\varepsilon>0$, and open $O$ : $X_{\varepsilon} \cap O$ has non empty interior.
Proof: WLOG we can assume that $O=(-1,1) \times (1,1)$. Since $x \mapsto f(x,0)$ is continuous at $0$, there exists $\alpha \in (0,1)$ such that $|f(x,0)-f(0,0)| \leq \frac{\varepsilon}{4}$ for all $|x| \leq \alpha$. Denote $$E_{\beta} = \Big \{ x \in (-\alpha,\alpha) ~|~ \forall |y| < \beta, ~ |f(x,y) - f(x,0)| \leq \frac{\varepsilon}{4} \Big\}.$$ Then : (1) $E_\beta$ is closed because it is an intersection of closed sets, (2) $\cup_\beta E_\beta = (-\alpha,\alpha)$ because $y \mapsto f(x,y)$ is continuous for any $x$. Baire theorem implies that $E_{\beta_0}$ contains an open $I$ for some $\beta_0$. Then for any $(x,y)$ and $(x',y') \in I \times (-\beta_0,\beta)$, we have $$\begin{array}{rcl} |f(x,y)-f(x',y')| &\leq& |f(x,y)-f(x,0)| + |f(x,0)-f(x',0)| + |f(x',y')-f(x',0)| \\ &\leq& \varepsilon \\ \end{array}$$ Hence $X_\varepsilon \cap O$ has non-empty interior. QED
Baire theorem implies that $\bigcap_{\varepsilon \in \mathbb{Q}_{>0}} X_{\varepsilon}$ is dense.
Solution 2:
This is definitely overkill but....
This article from The American Mathematical Monthly (Vol. 78, No. 2, Feb. 1971) shows that a function continuous in each variable is a Baire Class 1 function (the pointwise limit of a sequence of continuous functions). Then Baire's Category theorem applied to Theorem 2.2 of this shows that the function is actually continuous on a dense subset of the domain.
