First and foremost, it is important to know that open and closed are not opposites; i.e, a set that is not closed is not necessarily open. Sometimes sets can be neither open nor closed. For example, $[0,1)$. Sometimes sets can be both open and closed. For example, the emptyset or $\Bbb{R}$. One way to define an open set on the real number line is as follows:
$S \subset \Bbb{R}$ is open iff for all $s \in S$, there exists an interval of the form $(a,b)$ such that $s\in(a,b) \subset S$.
Another way to tell if a set is open is if it is the complement of a closed set. If $C$ is a closed set, then $\Bbb{R} \setminus C$ is open. Let's consider the union of open sets $(-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty)$. This union is open (although you should prove that any union of open sets is open so you can know this). Now, the complement is $$\Bbb{R} \setminus [(-\infty,1)\cup(1,2)\cup(2,3)\cup(3,\infty)]= \{1,2,3 \}$$ so we now see that the complement of $\{1,2,3 \}$ is open, allowing us to deduce that $\{1,2,3 \}$ is closed. Read the definitions carefully of open sets, closed sets, limit points and boundary points. A clear understanding of the differences and how they interact will take you far in real analysis and topology.
"Opposite" as a term will only mislead you. The entire space $\mathbb{R}$ is both open and closed, as is the empty set $\emptyset$; neither is the "opposite" of itself. The complement of an open set is closed, and vice versa.
If $E$ is not open then it is closed, and vice-versa.
This is false. Counterexample: $[0,1)$.
if $E$ is not open and $E^c$ is not closed then $E$ is closed and $E^c$ is open.
False. Saying that $E^c$ is not closed is equivalent to saying that $E$ is not open, so the hypothesis is just "$E$ is not open". From this, as noted in 1., it doesn't follow that $E$ is closed – as mentioned, they are not "opposites".
Let $E^o$ be the set of all interior points of a set $E$. Prove $E^o$ is always open.
$x$ is an interior point of $E$ iff some open neighborhood $U$ of $x$ is contained in $E$. So if $y \in E^o$, then $y \in U \subseteq E$ for some open neighborhood $U$ of a point $x \in E$. But then $y$ is by definition an interior point of $E$, as this same $U$ is an open neighborhood of $y$ contained in $E$.
Best Answer
Neither. It doesn't contain a neighbourhood of $1$, so it isn't open; nor is its complement, $(-\infty,\,0]\cup (1,\,\infty)$, which doesn't contain a neighbourhood of $0$.