Closed Intervals How do topologists prove continuity of a function with the usual topology at the endpoints of a closed interval?  For instance, how would a topologist prove continuity for $f(x)=x^2$ on the closed interval $[0,1]$ at the point $x = 1$ using open sets (not using calculus)?

Same way you would do it on an open interval. You have to show that inverse images of open sets are open, for $f(x)=x$ the inverse image of any set is the set itself, so it's true. It is only slightly harder for $f(x)=x^2$, but the inverse image of $(a,1]$ say, is $(\sqrt{a},1]$ and both are open (it's enough to consider intervals containing $1$ for continuity at $1$).
I think your issue is that open neighborhoods of $1$ do not "look" open on $[0,1]$ in the usual sense, but intersection of any open set in $\mathbb{R}$ with $[0,1]$ is nonetheless open in $[0,1]$ by definition of induced topology on $[0,1]$. So for example, $(\frac1n,1]$ are all open for any $n\in\mathbb{N}$.
