Need help understanding inferring proposition is false from removing out of an inconsistent set

I have been reading Introduction to formal logic by Peter Smith. In his Exercise 2 he has a question and answer as follows These questions are either true or false. I cannot seem to wrap my head around this answer.

Question) If a set of propositions is inconsistent, then if we remove some proposition P, we can validly infer that P is false from the remaining propositions in the set.

Answer) True. Suppose some set S of propositions plus P is inconsistent. That means there is no way of making those propositions in S true and making P true too. That means that any way of making the propositions in S true must make P false. Which is to say that we can validly infer that P is false from the propositions in S.

I'm confused because to me this stating "some proposition P" means picking an arbitrary proposition out of our inconsistent set. If this is arbitrary then how could we know we didn't simply grab a true proposition from our set? His answer seems to me that we hand pick the false proposition out of our set. I really don't even understand how with only knowing consistency of our set, which has nothing to do with the truth/falseness of the propositions, we can conclude the truth/false of a given proposition within the set. Could someone attempt to elaborate or answer this in a different way? Possibly give an example with an actual set of inconsistent propositions. I'm new to logic so I may be overlooking something stupid simple.

From a bit more syntactic perspective, there are two cases. Let $S$ be a set of propositions not including $P$ such that $S\cup \{P\}$ is inconsistent. In the first case, $S$ is inconsistent on its own. In this case, we can prove anything including $\neg P$. In the second case, $S$ is consistent, and it's only when we add $P$ that we get a contradiction. But proving that assuming $P$ leads to a contradiction is exactly what it means to prove $\neg P$, so we necessarily can prove $\neg P$ from $S$. Thus, in either case we can prove $\neg P$.
We didn't actually need to break the argument into two cases. The "second case" actually is a full argument by itself (dropping the assumption that $S$ is consistent). I decided to break it into two cases to help emphasize why the "truth" (provability) of $P$ is not important. In more semantically-oriented language, if $P$ is "true", then $S$ must be "false" corresponding to the $S$ is inconsistent case. And from the semantic perspective, "false" implies "true" and "false" implies "false" are both "true".
• The remaining propositions being an inconsistent set is the first case I mentioned, though, as I also mentioned, the argument for the second case doesn't actually rely on $S$ being consistent. Commented Jun 8, 2018 at 7:27