Every infinite recognizable language has an infinite decidable subset. $($Hint: Use the equivalences between recognizers and enumerators, and deciders and ordered enumerators$).$

Solution:

Idea: Let $L$ be any infinite recognizable language. Let $E$ be an enumerator for it$.$ We will modify $E$ to only enumerate strings in lexicographic order. This modification will generate an infinite decidable subset of $L.$

Modification: Add two extra tapes $T_1$ and $T_2$ to $E.$ On $T_1$ it will work similar to its main tape, and on $T_2$ it will record the latest accepted string. Whenever it reaches accept state of the original enumerator, it compares $T_1$ with $T_2$ to check if $T_1$ has a string $x$ greater than the string $y$ on $T_2.$ If no$,$ then it keeps enumerating on $T_1$ as usual. If yes, then it writes $x$ on $T_2$ and on the main tape, goes to accept state to accept $x,$ and then continues enumerating on $T_1$ as usual.

Proof: Modified E$\text{'}$s language satisfies following three properties because:

$($a$)$ Subset of $L(E)(0\mathrm{.}5$ points$)$:

$($b$)$ Decidable $(1$ point$)$:

$($c$)$ Infinite $(0\mathrm{.}5$ points$)$: