subsets of infinite recursive sets
It is correct. 

Every infinite set has an undecidable subset, you can use the cardinality argument: $\aleph_0\leq C \implies \aleph_0 &lt; 2^C$. In fact most of its subsets are undecidable (and you can replace undecidable with any countable class of languages, e.g. c.e., arithmetical, analytical, ...).

The bad thing about this argument is that it doesn't give any information about how difficult the subset is. We usually want a subset which is as easy as possible. One way to get this is using a diagonalization similar to the cardinality argument using the fact that $C$ is decidable:

Define $D = \{ i \in C \mid i \notin W_i \}$, where $W_i$ is the $i$th c.e. set. Obviously $D \subseteq C$. Moreover $D$ can be solved with an oracle for $C$ and $K = \{ i \mid i\notin W_i\}$. So if $C$ is decidable then $D$ is a co-c.e. language.