Connectedness arguments in elementary mathematics? To begin, let me explain a proof strategy (which I'll call the connectedness principle for want of a better, more canonical term):
One way to prove that a mathematical object $O_1$ has some property $P$ is to: 


*

*Construct a (topological) space $\mathfrak M$ containing the object $O_1$ such that $P$ is a continuous invariant in $\mathfrak M$, meaning that if $t \mapsto O_t$ is a continuous path of elements of $\mathfrak M$, every element $O_t$ has property $P$ as soon as one of them has it.

*Prove that some element $O_0$ in $\mathfrak M$ has property $P$.

*Prove that $O_0$ and $O_1$ are linked by a path $t \mapsto O_t$ in $\mathfrak M$ (for example, prove that $\mathfrak M$ is path-connected).


This strategy is for example well illustrated by a very nice proof of the genus-degree formula.
My question is do you know any example of an application of this strategy in more elementary mathematics? 
After all, even in high school or in the first years of college, the mathematical world is full of elements living in connected spaces (points, lines, triangles, numbers, functions...) and I see no reason that this proof strategy couldn't work in this context, even without the topology jargon.
Of course, one can imagine variants of this strategy (restraining oneself to polygonal paths, for instance) and I'm interested in all of these. However, I would like the answers to keep a topological flavour: I'm not interested (in this question) by examples using other kinds of invariance (e.g. through the action of a group).
A last remark: I haven't found this strategy in problem-solving books so I don't know if it has a well-established name. If you know of such a name, or of a book mentioning this strategy, please tell me!
 A: I think you've misrepresented the first step in your proof strategy, because it's too strict and as such short-circuits all relationship to continuity (and you'll notice that your link does not use this method). Specifically, rather than proving that any continuous path preserves the property, you need only prove that the property is clopen, i.e. for any $x$ which satisfies $P(x)$ there is a neighborhood of $x$ which also satisfies it, and similarly for $\neg P(x)$. Furthermore, you need not have $\frak M$ path-connected, but only connected, because otherwise $\{x|P(x)\}$ and $\{x|\neg P(x)\}$ form a disconnection of the space.
This might be a bit too trivial, but the example that comes to mind for me is the proof of the intermediate value theorem. Let ${\frak M}=[0,1]$, $O_1=1$, $O_0=0$, and $$P(x)\iff f(x)<0\vee\exists y:f(y)=0.$$ (Note that the second disjunct does not depend on $x$.) Then if $f(x)$ is a continuous function on $[0,1]$ with $f(0)<0$ and $f(1)\ge 0$, we have $P(0)$ satisfied, and for each $x$, either $f(x)<0$ and then there is an open set $U$ around $x$ such that $f(y)<0$ for all $y\in U$, i.e. $P(x)$ is true on $U$, or the second disjunct is true so that $P(y)$ for all $y$. Otherwise $f(x)>0$ and the second disjunct is false, so $P(y)\iff f(y)<0$ for all $y$, which is false in a neighborhood of $x$. Thus the connectedness argument gives that $P(x)$ is true for all $x\in [0,1]$, and since $f(1)\ge0$ the first disjunct is false, so $\exists y:f(y)=0$.
A: Well you can use it to prove that two homotopic contours in the plane minus a point have the same winding number about the point. This is done by integrating the function 1/z around the contours. Maybe this is not elementary enough. 
