A question about Baby Rudin Theorem 2.27 (a)

Solution 1:

$p$ was chosen randomly, so could be any point not in $E$, not in $E'$, as any such $p$ was explicitly ruled out as a limit point of $E$, hence the existence of a neighborhood of $p$ will not intersect $E'$.

Your proof is fine. You explained nicely why Rudin's claim follows.

Rudin is notorious for leaving some of the "links" between "stepping stones" (steps in his proofs) "to the reader". You've just filled in some of those unwritten details, and you are correct in those details. It's always a good idea to do so as you read any text, when anything is not immediately apparent to you while reading through a proof. In particular, that can often be the case when reading Rudin. When you revisit the proofs, then, that "extra work" will pay off, as you'll have made the connections, and will then be able to reread Rudin's proofs and follow them without so much effort.

Sometimes filling in the details may amount to simply "unpacking" the definition(s) of the terms being used in a theorem (what it means to be a limit point, e.g.). You'll find that's a sound way to learn the definitions inside-out, and to review theorems when they are used in subsequent proofs. (In all honesty, I personally "wrote, expanded upon, rewrote, extended, rewrote again" virtually all of Baby Rudin when I first encountered the text.

Solution 2:

To amplify slightly on @ncmathsadist's terse comment: It is a matter of negating a definition. If $p$ is not a limit point, then some neighborhood of $p$ must fail to intersect $E$. But then every point $q$ in this neighborhood has that same neighborhood that fails to intersect $E$, so $q\notin \bar E$.

P.S. Rudin is an ambitious choice for self-study. I continue to [complain] that he draws not a single picture, presumably as a matter of pride. Draw lots of pictures!