 Solution

Every word in $C_n(r)$ has probability at least $2^{-nr}$, so normalization gives $|C_n(r)|\le 2^{nr}$. Choose $H(U)<r<R$ and then $\epsilon <r-H(U)$. Every $\epsilon$-typical word satisfies $p(u^n)\ge 2^{-n(H(U)+\epsilon )}\ge 2^{-nr}$, hence $T_{\epsilon }^{(n)}\subseteq C_n(r)$ and $\mathrm{Pr}[C_n(r)]\to 1$. Injectively encoding $C_n(r)$ and using a default codeword outside it gives rate at most $R$ and vanishing error.