Is the statement $A \in A$ true or false?

It seems that you confused $A\in B$ ($A$ is an element of $B$) and $A\subset B$ ($A$ is a subset of $B$, that is, every element of $A$ is an element of $B$).

Edit: Example: $1\in \{1,2\}$, and $\{1\}\subset \{1,2\}$.

Every set is a subset of itself (this is what you argue in 1), but a set can never be an element of itself (at least in standard set theory, that is, Zermelo-Fraenkel set theory, where the axiom of regularity forbids this).

First let us get things in order.

The $\in$ symbol is for membership, whereas $\subseteq$ is for being a subset. Therefore $x\in A$ if $x$ is an element of $A$; whereas $B\subseteq A$ if every element of $B$ is also an element of $A$.

Reading your question again shows the discrepancy between the mathematical notation and the words explaining it. While $A\in A$ is not necessarily true, $A\subseteq A$ is always true.

Returning to the question whether or not it is true or false that $A$ is an element of itself is a vastly more complicated question because it cannot be discussed in a naive set theoretical setting. In the usual axiomatization of set theory, i.e. Zermelo-Fraenkel with Choice (ZFC), we include the axiom of regularity (known as axiom of foundation as well) which ensures that $A\notin A$ for any set.

However there are merits for using non-well founded set theory as well, in such set theory we do have that for some sets it is possible to have $A\in A$, but for example $\varnothing\notin\varnothing$ even in such settings. There have been uses for sets of the form $x=\{x\}$ in the past, however these uses often end up having reasonable equivalents in well-founded set theories (i.e. set theories which include the axiom of regularity).

So for most mathematicians the question is it true that $A\in A$ is always false, because working in ZFC (or an extension thereof) we never have this sort of situation.

Some people, however, do work in theories such as New Foundations which you have mentioned, and in such theories it is true for some sets that they are members of themselves, e.g. the universal set $V$ has this property that $V\in V$. However it is still true that $\varnothing\notin\varnothing$, so we cannot prove that every set is a member of itself. We cannot prove that every non-empty set is a member of itself since for $A=\{\varnothing\}$ we have that $A$ is non-empty (it has one element) but $A\notin A$ since the only element of $A$ is empty and $A$ is not empty.