Is every subset of a decidable set, also decidable?
This is a common misconception: complexity is not a measure of size. That is, it's not that "bigger" language are harder.
Intuitively, a language becomes harder when it's harder to describe it (TMs being a form of description). For example, as @Yuval Filmus points out in the comments, the language whose description is "everything" is very easy to decide. 

Similarly, the converse is not true - that is, smaller languages are not "harder" as well. For example, the language "nothing" is also easily decidable.

So the containment relation does not preserve hardness. Indeed - that's why we use the relatively complicated notion of reductions between languages, rather than showing containment.

So the simple answer to your question is that it's false, and an example, as in the comments, is $\Sigma^*$, which is decidable, but contains an undecidable language (indeed, every undecidable language).