A topological space $X$ has fixed point property if for every continuous map $f:X\to X$ there exists $x\in X$ such that $f(x) = x$. Its easy to see that fixed point property is a topological property, so that its preserved by homeomorphisms.
If $X$ has fixed point property, then it has to be connected, since if $X$ is not connected we can find open non-empty disjoint sets $U, V\subseteq X$ such that $U\cup V = X$, and taking $x\in U, y\in V$ and letting $f(z) = \begin{cases} y & z\in U\\ x & z\in V\end{cases}$ we see that $f$ is a continuous map without a fixed point.
But it doesn't have to be path-connected, since topologist's sine curve is an example of a space with fixed point property that's not path-connected.
Most examples of spaces with fixed point property are spaces like $[0, 1]^n$, topologist's sine curve or Warsaw circle. All of those spaces are compact.
Is every space with fixed point property also compact? If not, can we obtain any partial results that make this true?
The answer in general is no. The space $X = [(0, 0), (1, 0)]\cup \bigcup_{n=1}^\infty [(1/n, 0), (1/n, 1)]\subseteq\mathbb{R}^2$, where $[a, b]$ is the segment from $a$ to $b$, is an example of a non-compact subset of the plane with fixed point property (see example 3 in [1] for proof). Note that all subsets of $\mathbb{R}$ with fixed point property are compact, so the example is optimal in terms of embeddability in $\mathbb{R}^n$.
Convex sets
However, having fixed point property does imply compactness in some special cases. Schauder-Tychonoff fixed point theorem says that any non-empty compact convex subset of a locally convex topological vector space has the fixed point property (also see [2]).
Klee proved that a non-compact convex subset $C$ of a locally convex metrizable topological vector space $E$ must contain a closed copy of $[0, 1)$, so that $C$ lacks fixed point property (see [3]).
Thus for non-empty convex subset $C$ of locally convex metrizable topological vector space we have an equivalence: $$C\text{ has fixed point property iff }C\text{ is compact.}$$
Note however that there exists a locally convex topological vector space with a non-compact convex subset that has the fixed point property. If we let $E$ to be $C(\omega_1+1)^*$ with $w^*$-topology and $C$ be the set of all probability measures in $E$ supported in $\omega_1$, then $C$ is a convex non-compact set with fixed point property. See [4].
Manifolds and Peano continua
In [1], theorem 3, they prove that any locally connected, locally compact metrizable space with fixed point property is compact. See also theorem 2.7 in [3]. In particular any such space is a Peano continuum.
This also implies any manifold (with boundary) with fixed point property is compact. The converse is not true since $S^1$ doesn't have fixed point property.
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