We know that a function $f: R^2\to R$ that is section-continuous (that is each $x\mapsto f(x,y)$ and $ y\mapsto f(x,y)$ are continuous) need not be continuous. $f(x,y)=\frac{xy}{x^2+y^2}\chi_{\{0\}^c}$ is a counterexample for such a claim. However apparently if add the condition that $f$ maps compact sets to compact sets, then $f$ is continuous.
Note the the converse is always true, that is continuous maps send compact sets to compact sets.
I was wondering why that original condition is true? that is section continuous functions that send compact sets to compact sets must be continuous.
A proof by contradiction: suppose that $f$ satisfies the conditions but it is not continous; w.l.o.g. we may assume that $f$ is not continuous at $(0,0)$ and $f(0,0)=0$. Then there exists some $\varepsilon>0$ and a sequence $a_1,a_2,\ldots$ of points such that $a_n\to(0,0)$ and $|f(a_n)|\ge\varepsilon$.
Each point $a_n$ can be connected with the origin by a horizontal and a vertical line segments; on this set $f$ is continuous; so there is some point $b_n$ on those segments such that $|f(b_n)|=(1-\frac1n)\varepsilon$. Hence, $b_n\to(0,0)$, so the set $\{(0,0),b_1,b_2,b_3,\ldots\}$ is compact, but its image is not, contradiction.