CFG that generates $1^*$ is decidable
Maybe you're confusing two different problems.
The algorithm you are describing shows that the problem of testing whether a CFG generates some string from $1^∗$ is decidable (e.g., see here at page 21). Anyway, the problem of testing whether a CFG generates all the strings of the language $1^∗$ is truly decidable. Consider all production rules to be simple (I mean, no &quot;$|$&quot; on the right), and eliminate all productions including a teminal character different from 1 on the right. Now you have a CFG over a terminal alphabet of a single letter, and then it generates a regular language (see here). All that remains is to test if two regular languages are equal. Moreover, there are effective procedures to convert a CFG over a one letter (terminal) alphabet into a regular grammar (e.g., see here).

I was forgetting the last part of your question: the fact that it is undecidable whether $R\subseteq G$, where $R$ is a regular language, and $G$ is a CFG, does not implies that for some specific regular languages you can not decide the inclusion. For example, if $R$ is finite (possibly empty), then testing for $R\subseteq G$ is decidable (see the first link above).