Examples of arguments from connectedness Suppose $X$ is a connected topological space. A typical way that we prove a property $P(x)$ holds for all $x \in X$ is to show that $P$ is an open and a closed condition, and that $P(x)$ for some $x \in X$. 
Most recently I saw this show up in the proof that a connected, locally path connected space is path connected. (Pick a point $x \in X$ and let $P(y)$ be "There exists a path from $y$ to $x$".)
This seems to also be the idea behind some proofs that $[a,b]$ is compact. See here. 
I wanted to accrue some more examples of where we use this technique (for teaching purposes, perhaps). Does anyone have some more?
 A: Uniqueness of lifting: Given a covering space $p \colon X' \to X$, a pointed function $f \colon (Y,y_0) \to (X,x_0)$ where $Y$ is connected and a value $x' \in p^{-1}(x_0)$, there exists at most one lift $f': Y \to X'$ such that $f = pf'$ and $f'(y_0)=x'$.
The proof uses the property $P(x)$: $f'_1(x)=f'_2(x)$; it is open, closed and true for $y_0$.
A: One of my favorites is the proof that any connected manifold $M$ is homogeneous.  The steps are:


*

*Prove that, for any two points $p$ and $q$ in the interior of a closed $n$-ball $B^n$, there's a homeomorphism $B^n\to B^n$ that maps $p$ to $q$ and is the identity on the boundary.  (Just map straight line segments emanating from $p$ to straight line segments emanating from $q$.)

*Deduce that, for any point $p\in M$, the set of points to which $p$ can be mapped by a homeomorphism of $M$ is open.

*Use the same argument on the complement to show that the set of points to which $p$ can be mapped by a homeomorphism of $M$ is closed.

*Conclude that $M$ is homogeneous.
In fact, the same proof shows that, for a manifold $M$ with boundary whose interior is connected, there's a homeomorphism mapping any interior point $p$ to any other interior point $q$ that restricts to the identity on the boundary.  It follows easily that any connected manifold $M$ of dimension $\geq 2$ is $n$-homogeneous for all $n$, i.e. for any points $p_1,\ldots,p_n,q_1,\ldots,q_n \in M$ there exists a homeomorphism $M\to M$ mapping $p_i$ to $q_i$ for each $i$.
A: Uniqueness of analytic continuation: let $\Omega$ be a connected open subset of $\mathbf{C}$, and let $u$ analytic on $\Omega$. If $\{z\in\Omega \ | \ u^{(k)}(z)=0 \ \mathrm{for \ all} \ k\geqslant 0\}$ is nonempty, then $u=0$. (The set in question is both open and closed, thus must be all of $\Omega$ by assumption.)
Another example coming from analysis: let $\Omega$ be a connected open subset of $\mathbf{C}$, and let $f\in C^0(\Omega)$ with $$f(x)=\frac{1}{2\pi}\int_0^{2\pi} f(x+r\exp(i\theta))dx$$
for all balls $B(x;r)\subset\Omega$. If $f$ is bounded and achieves its maximum, then it must be constant. (The set $\{x \ | \ f(x)=\sup_\Omega f\}$ is both open and closed.)
A: I know of at least two ways in which connectedness is commonly used:


*

*To show that a connected topological space has property $X$, you show that $X$ is both a closed condition (i.e. closed under taking limits) and an open condition (i.e. can be extended locally). Some of the other answers here use this technique, and I've seen it applied in PDEs, Riemannian geometry, etc.

*To show that a map from a connected space is trivial, given that its image is discrete.
This for instance shows that winding numbers for curves in $\mathbb{C}$ are constant over connected regions.
Here's a sketch of the standard proof that connected compact groups are unimodular (i.e. have left and right invariant Haar measure).
Let $G$ be a compact group, $\omega$ a left invariant volume  measure on $G$. Then for any $g \in G$, $f \mapsto \int_G f(xg^{-1}) d\mu(x)$ is also a left invariant Haar measure, so by uniqueness must be a scalar multiple of $\mu$. In other words, $R_g^{-1}d\mu$ = $\phi(g)d\mu$. It is easy to see that $\phi$ is a continuous homomorphism from $G$ to $\mathbb{R}$, and since $G$ is compact, its image must be a compact group. This leaves two possibilities: $\{1\}$ or $\{\pm 1\}$, so by connectedness it must be the former.
You can also use this to show that group actions by connected groups on manifolds are orientation-preserving (because the identity is).
A: The proof of the Calabi conjecture. This is much beyond the other examples here in level, and is also the most illustrious one.
A: A nice one I saw recently in Warner's Foundations of Differentiable Manifolds and Lie Groups, in defining orientation on a manifold:
Suppose $M$ is connected. Consider $\bigwedge^{\text{top}}TM \setminus Z$, where $Z$ is the zero section of top-degree forms. Each fiber of $\bigwedge^{\text{top}}TM \setminus Z$ has exactly two components, so a connectedness argument (and local triviality of the bundle) shows that $\bigwedge^{\text{top}}TM \setminus Z$ must have at most two components. If it has one, the manifold is non-orientable; if it has two, $M$ is orientable.
