= Solution
{c}
\b[False], with “<disjoint union>” understood as a <topological disjoint union>. Take the <rational numbers> $\mathbb Q$ with the usual <subspace topology> from the <real line>.
Every <connected subset> of $\mathbb Q$ has at most one point. Indeed, if it contained $r<s$, choose an <irrational number> $t\in(r,s)$. The two relatively <open sets>
$$
\mathbb Q\cap(-\infty,t),\qquad \mathbb Q\cap(t,\infty)
$$
partition $\mathbb Q$ and separate any <subset> containing both $r$ and $s$. Hence the <connected components> of $\mathbb Q$ are <singletons>.
If a <topological disjoint union> of nonempty <connected spaces> were <homeomorphic> to $\mathbb Q$, the <image> of each summand would be <connected>, hence a <singleton>. Every summand would therefore be a one-point <topological space>. Their <topological disjoint union> is a <discrete space>, since each summand is <open>. But $\mathbb Q$ is not a <discrete space>: for each $r\in\mathbb Q$, the distinct <rational numbers> $r+1/n$ approach $r$, so $r$ is not an <isolated point>. This is a contradiction. The <set> decomposition into <connected components> always exists; giving those pieces the <topological disjoint union> topology need not recover the original <topology>.
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