Is every subset of a RE language also RE, in general? I'm trying to understand the question in my title in an intuitive way: 
If I have an RE language A, then some TM, say TM(A) accepts on it. If I take a subset of A, say A2, then all elements of A2 will cause TM(A) to halt in the accept state.

However, is it in general possible to then create a TM for A2, say TM(A2), such that A2 is the max possible set that halts TM(A2) in the accept state. Thus making A2 its language - and would this language be RE?

Thanks
No, it is not true that every subset of a recursively enumerable (RE) language is also RE. In fact, it make sense to say that a subset of a RE language is not RE more often than not.

For an example, let us consider the RE language, $\Sigma^*$ that contains all words. It is the language of the Turing machine which always halts and accepts in its first step. Note that every language is a subset of $\Sigma^*$. Every language that is not RE is an example.



Here are a few exercises for you to check.

Exercise 1. Is every subset of a regular language regular?

Exercise 2. Is every subset of a context-free language context-free?

Exercise 3. Is every subset of a recursive language recursive?