$f$ is continuous in $x$ and $y$ and maps compact set into compact set. Show that $f$ is a continuous function on $\mathbb R^2$. 
Suppose $f$ is defined on $\mathbb{R}^2$, $f$ is continuous in $x$ and $y$ respectively and $f$ maps compact set into compact set. Show that $f$ is a continuous function on $\mathbb R^2$.

Suppose $f$ is not continuous at $(x_0,\,y_0)$, then there are $\varepsilon_0>0$ and a sequence $\{(x_n,\,y_n)\}$ converging to $(x_0,\,y_0)$ such that 
$$
|f(x_n,\,y_n)-f(x_0,\,y_0)|\geq\varepsilon_0.
$$
Let $K=\{(x_n,\,y_n)\}_{n=0}^\infty$. $K$ is compact. Thus, $f(K)$ is also compact. Then I don't know what to do and am unable to use the condition that $f$ is continuous in $x$ and $y$ respectively. 
 A: This is quite an interesting question.
  From the point where you are stuck, the main insight is that one can construct using
  the assumption of separate continuity a new sequence which then yields a contradiction
  to the condition that $f(K)$ is always compact if $K$ is.
  Let us dive into the details.
Step 1:
  I claim that if $Q \subset \mathbb{R}^2$ is a rectangle, then $f(Q)$ is an interval.
  This follows more or less easily by noting that in a rectangle, you can always connect two
  points by a path which consists of two parts: in the first part, only the first coordinate
  changes, and in the second part only the second coordinate changes.
  One then uses the separate continuity of $f$ and the intermediate value theorem.
For completeness, here is a formal proof.
  By definition of an interval, what we need to show is that if $y = f(\xi), z = f(\eta) \in f(Q)$,
  and if $y < t < z$, then also $t \in f(Q)$.
  To see this, write $\xi = (\xi_1,\xi_2), \eta = (\eta_1,\eta_2)$, and consider the paths
  $\varphi : [0,1] \to Q, t \mapsto \big( (1-t) \xi_1 + t \, \eta_1, \xi_2 \big)$ and
  $\psi : [0,1] \to Q, t \mapsto \big( \eta_1, (1-t) \xi_2 + t \, \eta_2 \big)$,
  which are both well-defined (they really take values in $Q$) since $Q$ is a rectangle.
  Now, since $f$ is separately continuous, it follows that $f \circ \varphi$ and $f \circ \psi$
  are continuous.
  By the intermediate value theorem, it follows that $f(\varphi([0,1]))$
  and $f(\psi([0,1]))$ are both intervals, which intersect since $\varphi(1) = \psi(0)$.
  Therefore, their union is again an interval.
  We have thus shown $f(\xi), f(\eta) \in I := f(\varphi([0,1])) \cup f(\psi([0,1])) \subset f(Q)$,
  from which we get $t \in I \subset f(Q)$, since $I$ is an interval.
Step 2:
  Assume that $f$ is not continuous at $x_0 \in \mathbb{R}$, so that there is $\epsilon_0 > 0$
  and a sequence $(x_n)_{n \in \mathbb{N}}$ such that $x_n \to x_0$, but $|f(x_n) - f(x_0)| > \epsilon_0$
  for all $n \in \mathbb{N}$.
  By taking a subsequence, we can assume that either $f(x_n) \geq f(x_0) + \epsilon_0$ for all $n$,
  or that $f(x_n) \leq f(x_0) - \epsilon_0$ for all $n$.
  For simplicity, I only consider the first case in what follows.
Note that $x_n \neq x_0$ and hence $\epsilon_n := \| x_n - x_0 \|_{\ell^\infty} > 0$ for all $n$.
  Define $Q_n := x_0 + [-\epsilon_n, \epsilon_n]^2$ for all $n \in \mathbb{N}$, noting that $Q_n$ is a rectangle
  with $x_n, x_0 \in Q_n$.
  By Step 1, this implies that $f(Q_n) \supset [f(x_0), f(x_0) + \epsilon_0]$ for all $n \in \mathbb{N}$.
  We can therefore choose $y_n \in Q_n$ satisfying $f(y_n) = f(x_0) + \frac{\epsilon_0}{2} + \frac{1}{n}$,
  at least for $n$ so large that $n^{-1} < \epsilon_0 / 2$.
Because of $y_n \in Q_n = x_0 + [-\epsilon_n, \epsilon_n]^2$ and $\epsilon_n \to 0$,
  we have $y_n \to x_0$, so that $K := \{ y_n \colon n \in \mathbb{N} \} \cup \{x_0\}$
  is compact.
  But by construction, we have
  $$
    f(K)
    = \Big\{ f(x_0) + \frac{\epsilon_0}{2} + \frac{1}{n} \colon n \in \mathbb{N} \Big\} \cup \{ f(x_0) \},
  $$
  from which it is easy to see that $f(K)$ is not compact.
  This is the desired contradiction.
