Continuity of a function in two variables 
Function $f(x,y)$ is continuous in each variable separately. Prove that there exists a point where it is continuous in two variables.

I do not quite understand how to act here. I know the definition of continuity. But how to use them, I do not know. This is my homework on mathematical analysis.
 A: 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.
A: 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. 

A: Let $(x_p, y_p)$ be a point and let ${(x_n , y_n)}_{n\in \mathbb{N}}$ be any sequence such that $(x_n , y_n) \rightarrow (x_p, y_p)$.
Then
$$
\lim_{n\rightarrow \infty} |f(x_n , y_n) - f(x_p , y_p)| = \lim_{n\rightarrow \infty} |f(x_n , y_n) - f(x_n , y_p) + f(x_n , y_p) - f(x_p , y_p) + f(x_p , y_p) - f(x_p , y_n) + f(x_p , y_n) - f(x_p , y_p)| \leq \lim_{n\rightarrow \infty} |f(x_n , y_n) - f(x_n , y_p)| + |f(x_n , y_p) - f(x_p , y_p)| + |f(x_p , y_p) - f(x_p , y_n)| + |f(x_p , y_n) - f(x_p , y_p)| 
$$
Each of these terms goes to zero since $f$ is continuous in each variable. 
Therefore $$\lim_{n\rightarrow \infty} f(x_n,y_n) = f(x_p,y_p)$$ By definition this means f is continuous at $(x_p, y_p)$.
(This proof is straight forward to generalize in arbitrary metric spaces, though I figured that's not what you were looking for.)
