TITLE: Every infinite regular language has a non-regular subset?
QUESTION [5 upvotes]: I found some examples of infinite regular languages having non-regular subsets. Is it true in general?

REPLY [2 votes]: Let $L$ be any infinite language.  Because the number of subsets of $L$ is uncountable, while the number of regular languages is countable, there must be subsets of $L$ that are not regular.  The same reasoning shows that there are subsets of $L$ that are not context-free, that are not recursive (decidable), and that are not recursively enumerable (recognizable), since all of these families of languages are countable.
If an explicit example is required, it is straightforward to construct a subset of $L$ that violates the pumping lemma for context-free languages, and hence that is not context-free (or regular).  We start with two empty sets, $A_{1}=B_{1}=\varnothing$.  We then repeat the following steps for $i\in \mathbb{N}$:

Choose $s \in L \cap B_{i}^{c}$ that is longer than the longest string in $A_{i}$.
Let $A_{i+1}=A_{i} \cup \{s \}$.
Let $B_{i+1}=B_{i} \cup \{ uv^2 x y^2 z : s=uvxyz \wedge |vy|\ge 1\}$.

Because $L$ is infinite and $B_{i}$ is always finite, there is always a string available to choose in step 1.  The set $A_{\infty}=\bigcup_{i=1}^{\infty}A_{i}$ is a subset of $L$ that violates the pumping lemma: it contains arbitrarily long strings, none of which can be "pumped" without producing strings in the forbidden set $B_{\infty}=\bigcup_{i=1}^{\infty}B_{i}$, which is disjoint from $A_{\infty}$ by construction.  We conclude that $A_{\infty} \subseteq L$ is not context-free.