Yes, this is an issue.
Naively, this issue cannot be dealt with, and we'll get to that in a moment. But in 1917 mathematicians already noticed that "normal sets" do not contain themselves, and in fact have an even stronger property. Namely, there are no infinite decreasing chains in $\in$, so not only that $a\notin a$ it is also true that $a\notin b$ whenever $b\in a$, and that $a\notin c$ whenever for some $b\in a$ we have $c\in b$; and more generally there is no sequence $x_n$ such that $x_{n+1}\in x_n$ for all $n$.
This is exactly what the axiom of regularity came to formalize. It says that the membership relation is well-founded, which assuming the axiom of choice, is equivalent to saying that there are no decreasing chains. In particular $A\notin A$, for any set $A$.
But we know, nowadays, that it is consistent relative to the other axioms of modern set theory (read: $\sf ZFC$) that there are sets which include themselves, namely $x\in x$. We can even go as far as having $x=\{x\}$. You can even arrange for infinitely many sets of the form $x=\{x\}$.
This shows that naively we cannot prove nor disprove that sets which contain themselves exist. Because naive set theory has no formal axioms, and is usually taken as a subset of axioms which include very little from $\sf ZFC$ in terms of axioms, and certainly it does not include the axiom of regularity.
But it also tells us that we cannot point out at a set which includes itself, if we do not assume the axiom of regularity. Since these sets cannot be defined in a nontrivial way. They may exist and may not exist, depending on the universe of sets we are in. But we do know that in order to do naive set theory and even more, we can safely assume that this situation never occurs.
You may find Chardrand, Polimeni & Zhang's Mathematical Proofs: A Transition To Advanced Mathematics useful. They teach us the intro class from this book in my university, and it has a whole chapter on proofs in advanced calculus (or introductory analysis, if you like). I have not read that chapter thoroughly (we don't learn about analysis in the intro class), but I have taken a look at it, and it has all these "proof strategy" parts before actual proofs, and the book in general does not skip any steps (i.e., no "proof is easy"'s :) ). Another suggestion would be Fitzpatrick's Advanced Calculus. Again, this book is very thorough with the proofs. There are lots of exercises in both of the books as well.
As for my humble suggestion on how to understand the proofs, I find it quite beneficial to first convincing oneself that the theorem must be the case by coming up with different examples. Modify the hypotheses and see where the theorem really breaks down. This is a slow process at first but the examples/counterexamples start to come to you faster (in general) as you continue. Finally, hang in there :) .
As a final remark, I find the following (trivial) lemma quite useful in proving results related to sequences: A sequence converges to $x$ iff for any subsequence of it there is a further subsequence that converges to $x$. This may count as a useful argument.
Best Answer
Call the set $S$. By "into" they mean any map $f\colon S\rightarrow S$. By "onto" they mean a map $f\colon S\rightarrow S$ whose image is $S$ itself (i.e. $f(S)=S$). That is, each element in the codomain is mapped to by some element in the domain.