Does a compact connected complete linear order have the fixed point property? Would the same arguments used for showing $[0,1]$ has the fixed point property hold in this general case?  What could go wrong? 
EDIT: The fixed point property can be interpreted in two ways that I am now aware of:
i) A partially ordered set has the fixed point property if every increasing self-map has a fixed point.
ii) A linearly ordered topological space has the fixed point property if every continuous self-map has a fixed point.
My question has been answered positively in the case of (i). I am also interested in knowing what happens if we interpret FPP as in (ii).
 A: 
Theorem (Tarski): If $(X,\leq)$ is a complete linear order, then every weakly increasing function $f: X \rightarrow X$ has a fixed point.

You can prove this using the Principle of Ordered Induction see Theorem 4 here, which is the generalization of Real Induction (Proposition 1 of loc. cit.) to any Dedekind Complete linearly ordered set.
(Recall that a linear order is Dedekind complete if every nonempty subset which is bounded above has a least upper bound; equivalently, every nonempty subset which is bounded below has a greatest lower bound.  A linear order is complete if every subset has a least upper bound and a greatest lower bound: this is equivalent to Dedekind complete plus the existence of top and bottom elements.)
The result when $X = [a,b]$ (a closed bounded interval in $\mathbb{R}$) is proved using Real Induction in Theorem 6.22 of these notes.  That proof works verbatim for any complete linear order.
Other comments:

*

*In fairness I should mention also the more standard approach: show that $\operatorname{sup} \{x \in X \mid f(x) \geq x\}$ is a fixed point.  This is quite straightforward, maybe even more so than the above approach.


*If you equip a linear order with the order topology, then it is compact iff it is complete, and it is connected iff it is Dedekind complete and dense (if $x < z$ there is $y$ with $x < y< z$).  This is also pleasant to prove using ordered induction: see here.  So you are assuming more than you need to: complete implies compact, and connected is neither necessary nor sufficient.


*As Asaf points out, the result is a special case of the Knaster-Tarski Theorem: that theorem applies to a complete lattice rather than a complete linear order and asserts that the set of fixed points, in addition to being nonempty, forms a sublattice.


*The fixed-point property characterizes completeness in lattices:

Theorem (Anne C. Davis) Let $X$ be a lattice in which every weakly increasing function $f: X \rightarrow X$ has a fixed point.  Then $X$ is complete.

Actually Davis's Theorem is quite a bit harder to prove than the Knaster-Tarski Theorem.
Added: Niels Diepeveen in the comments points out that the "fixed point property" might be: "Does every continuous (for the order topology) function $f: X \rightarrow X$ have a fixed point?"  That's a different question.  The following is an incomplete answer.

Proposition: Let $X$ be a nonempty linearly ordered set endowed with the order topology.
a) If $X$ is complete and connected, then every continuous $f: X \rightarrow X$
has a fixed point.
b) If every continuous function $f: X \rightarrow X$ has a fixed point, then $X$ is connected.

Proof: a) Suppose that $X$ is complete but that there is a continuous $f: X \rightarrow X$ which does not have a fixed point.  We'll show that $X$ is not connected.  Let $X_1 = \{x \in X \mid f(x) < x\}$ and $X_2 = \{x \in X \mid f(x) > x\}$.  Then $X_1$ and $X_2$ are open, and -- since there are no fixed points --
$X_1 \cup X_2 = X$.  Moreover the bottom element $0$ lies in $X_2$ and the top element $1$ lies in $X_1$, so we get a separation of $X$.
b) If $X = X_1 \coprod X_2$ is a separation of $X$, let $x_i \in X_i$ for $i = 1,2$,
and define $f: X \rightarrow X$ by mapping every point in $X_1$ to $x_2$ and every point in $X_2$ to $x_1$.  $f$ is locally constant hence continuous, but has no fixed point.
What remains is to determine whether every continuous function $f: X \rightarrow X$
having a fixed point implies that $X$ is complete.
