Proving continuity on a plane for a function continuous when restricted to segments Suppose we have some map $F:I\times I \mapsto X$ for some topological space $X$. We know that $F$ is continuous on any line segment $L\subset I\times I$. I'm hoping to show that $F$ is continuous on the entirety of $I\times I$. This makes intuitive sense but I have no idea on how to formalize it.
This arose while I was trying to look for an alternate proof of Munkres' lemma 54.2, in which applying lemma 54.1 leads to the condition of the problem above.
EDIT: $I=[0,1]$ is the closed unit interval on the real line, and $I\times I$ is the closed unit square
 A: This is not true. Here's a counterexample.
Define $f\colon I\times I \to \mathbb R$ by
$$
f(x,y) = 
\begin{cases}
\dfrac{x^2 y}{x^4 + y^2}, &(x,y)\ne(0,0),\\
0, & (x,y)=(0,0).
\end{cases}
$$
This is continuous everywhere away from $(0,0)$, because it's a rational function with nonzero denominator. But it's not continuous at the origin, because the sequence $(x_n,y_n) = (1/n, 1/n^2)$ approaches $(0,0)$, but $f(x_n,y_n) = 1/2$, which does not approach $f(0,0)=0$.  
However, this function is continuous along every line segment contained in $I\times I$. On any line segment that does not contain the origin, it's continuous because it's a composition of continuous functions there. On a line segment that includes the origin, we have three possibilities:
\begin{align*}
f(x,0) &\equiv 0, &&\text{(horizontal segment)},\\
f(0,y) &\equiv 0, &&\text{(vertical segment)},\\
f(x,mx) &= \dfrac{mx}{x^2+m^2}, &&\text{(segment with slope $m\ne 0$)}.
\end{align*}
In each case, the restriction of $f$ to the line segment is continuous.
