= Solution
Since $|D_5|=10$, <Lagrange's theorem> permits proper subgroup orders only $1,2,5$. The complete list is
$$
\boxed{\{I\},\qquad\langle R\rangle=\{I,R,R^2,R^3,R^4\},\qquad\{I,R^jm\}\quad(j=0,1,2,3,4).}
$$
To show completeness, an order-two subgroup must be generated by its unique nonidentity element, necessarily one of the five reflections. An order-five subgroup is cyclic, since its order is prime; its generator cannot be a reflection, and every nonidentity rotation generates the same rotation subgroup. There is only one order-one subgroup.
Thus there are seven proper subgroups when the trivial subgroup is included, or six nontrivial proper subgroups. These are the prime-sided case of the <classification of subgroups of a dihedral group>.
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