The typical example of a connected space that is not path connected (the topologist's sine curve) has the fixed point property.

Let $X_1 = \{0\} \times [-1,1]$ and $X_2 = \{(x,\sin(1/x)) : 0<x \le 1\}$ and $X = X_1 \cup X_2$

If $f:X \to X$ is a continuous map, then since $X_1$ and $X_2$ are both path connected, the image of each one of them must be entirely contained in another of them.

If $f(X_1)\subset X_1$ then $f$ has a fixed point because the interval has the fixed point property. If $f(X_2) \subset X_1$ then $f(X) = f(cl(X_2)) \subset cl(f(X_2)) \subset X_1$ and in particular $f(X_1) \subset X_1 $ again $f$ has a fixed point.

So the only case that remains is that $f(X)\subset X_2$ And since $X$ is compact, its projection to the first coordinate is also compact so that it must be an interval $[a,b]$ with $a>0$ Thus $f(X)$ is contained in $S = \{(x, \sin (1/x)) : x\in [a,b]\}$ But $S$ is homeomorphic to a closed interval, so that it has the fixed point property, and the restriction of $f$ to $S$ is a continuous map $S \to S$ so that it has a fixed point.

This proof is due to Koro.