I understand the assumptions that have to be true about a property or set of properties in a Turing machine description for Rice's Theorem to apply.
But then what? If a set of Turing machines have an undecidable property, is the language itself necessarily undecidable? Or only if you can find a reduction from a known-undecidable machine is the language undecidable?
What does it really say about a language if it has an undecidable property? Are machines that recognize that language undecidable?
An undecidable property $\pi$ of Turing machines is the same as an undecidable language consisting of all encodings of Turing machines satisfying $\pi$. We identify the property with the language of encodings of Turing machines satisfying the property.
The fact that a language is undecidable just means that no Turing machine decides the language — this is the definition of undecidability. In particular, no machines recognize the language, assuming recognize means decide; a machine $T$ decides a language $L$ if $T$ terminates on all inputs, on inputs $x \in L$ returns $1$, and on inputs $x \notin L$ returns $0$. A machine cannot be undecidable, only a language can be undecidable.
You are talking about two things here. Consider a non-trivial property $P$ of languages. $P$ can be anything, for instance, does the language contain a string of prime length or not.
Now there are a set of Turing Machines whose languages have this property $P$. Let,
$$S = \{\langle M\rangle \mid\text{ Language of M satisfies }P\}\,.$$
Now, Rice's Theorem says that $S$ is undecidable. This means that you cannot build a decider that takes the description of a Turing Machine and tells whether the language of the input machine has a particular non trivial property. It does NOT say that for any $L$, ($L \in P $) $L$ is undecidable.
Consider the language $L = \{10\}$. This language $L$ is in $S$ (note that I am saying $L \in S$). Clearly, the language is decidable even though $S$ is not.
Hope that clears your doubt.
