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Suppose that $X$ is a metric space with finite diameter. Is it possible to have a countable collection of disjoint open sets $U_i$ with $\text{diam}(U_i) \geq r > 0$, for some fixed $r$? I'm not assuming the $U_i$ are connected, if that matters. I can't seem to find anything about lower bounds on unions of diameters.

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Yes, note that any subset of a metric space is a metric space, so consider the unit ball in an infinite dimensional normed space, e.g. $\ell^\infty(\Bbb Z)$. Then as the unit ball is not compact, we can in the usual way done in functional analysis provide an infinite set of open balls of the same radius with no finite sub-cover, even though the space's diameter is clearly $1$.

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    Say I'm on a residual subset of a finite dimensional compact Riemann manifold? – 2017-02-24
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    Or just on a compact manifold if there can be infinitely many disjoint sets with a fixed diameter? – 2017-02-24
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    @Analysis the closure of that bounded set is compact since you can embed into Euclidean space, so infinite disjoint balls would give rise to an infinite cover with no finite subcover. If you want arbitrary sets, you can of course use the axiom of choice on $[0,1]$ for example to get a bunch of crazily shaped sets which are disjoint via $\Bbb R/\Bbb Q$. – 2017-02-24
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    I realized that as soon as you typed it! Thanks for answering the question, it's been a long week! – 2017-02-24
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    I can't accept for a few minutes, but I will – 2017-02-24
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    @Analysis my pleasure, glad I could help. – 2017-02-24
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Let $I$ be an arbitrary index set (infinite, uncountable, whatever) and let $U_i\ (i\in I)$ be a family of pairwise disjoint $2$-element sets. Let $X=\bigcup_{i\in I}U_i$ and define a metric $d$ on $X$ by setting $d(x,y)=1$ whenever $x\ne y.$ Then $\operatorname{diam}X=1$ and $\operatorname{diam}(U_i)=1$ for each $i\in I.$