= Solution
\b[No.] Take
$$
B=[0,1]\times\{0\}\subset\mathbb R^2.
$$
It contains more than one point and has the <fixed-point property of a closed interval>, but it is not homeomorphic to the closed unit disc. Indeed, deleting any point $(t,0)$ with $0<t<1$ disconnects $B$, so such a point is a <cut point>. Deleting one point from a two-dimensional closed disc leaves a <path-connected space>. Since being a cut point is preserved by <homeomorphism>[homeomorphisms], the two spaces cannot be homeomorphic. This gives the required <counterexample>.
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