Non-circular definition of 'interior' points and 'open' sets A little background
I am by no means a college-educated mathematician. I am a 15-year-old American high school student currently enrolled in AP Calculus BC, which is the equivalent of Honors Calculus I at university. I am also currently enrolled in the International Baccalaureate Programme: next year, I will take IB Math Higher-Level 2, wherein I will compose an extended essay, which I plan on writing over set theory. Basically, I am sorry if this question seems rudimentary, but bear with me, please.
Explanation
I was reading a PDF on Lebesgue integration when I cam across a seemingly simple definition:

$x$ is an interior point on $S$ if and only if the the interval $\left( x-\epsilon ,x+\epsilon\right)\subseteq S$ where $\epsilon >0$

Initially, this made enough sense to me. For example, take a look at the diagram below:

Here, $\epsilon >0$ and $\left( x-\epsilon ,x+\epsilon\right)\subseteq S$ since $\left( 1/4,3/4\right)\subseteq\left[0,1\right]$, meaning that every single point on $\left( 1/4,3/4\right)$ is also on $\left[0,1\right]$. Therefore, $x$ (aka $1$) is an interior point of $S$. Logically, this seems to make sense; after all, $1/2$ is smack in the middle of $\left[ 0,1\right]$.
However, in my mind, this definition seems trivial, because (and this is where my reasoning has obviously led me astray) it appears that every point on $S$ is an interior point of $S$. To explain this, let's take a closer look at $\left[ 0,1\right]$:

Here, we have zoomed incredibly close in around $0$—so close, in fact, that we no longer have a line of infinite points but rather three neighboring points that have absolutely no space between them. In this case, $\epsilon$ is still greater than $0$, albeit by the smallest amount possible. Furthermore, since $\left( x-\epsilon ,x+\epsilon\right)$ is written with curvy parentheses and therefore excludes $x-\epsilon$ and $x+\epsilon$, the only point actually on $\left( x-\epsilon ,x+\epsilon\right)$ is $0$, meaning that $\left( x-\epsilon ,x+\epsilon\right) =\left\{ 0\right\}$. By definition, $x=0$, which also means that $\left\{x\right\}=\left( x-\epsilon ,x+\epsilon\right)$ and that, by continuation, $\left\{x\right\}\subseteq\left( x-\epsilon ,x+\epsilon\right)$. I imagine that this last step is where I am going wrong: I am left to conclude that if $\left\{ 0\right\} =\left( x-\epsilon ,x+\epsilon\right)$ and $\left\{ 0\right\}\subseteq\left[ 0,1\right]$ then $0$ is an interior point of $\left[ 0,1\right]$ by the $\left( x-\epsilon ,x+\epsilon\right)\subseteq S$ definition.
The more I thought about this, the more ridiculous it seemed. I could simply change the specific numbers in my previous two "proofs" to show that every point on an interval $S$, whether they have neighbors in $S$ or not, is an interior point of $S$.
The question

*

*Where did my reasoning go wrong?

*Could you please provide a working definition of an interior point that you believe I could understand?

*Would you also please expound upon that definition in order to distinguish between open and non-open sets. (According the afore mentioned PDF, an "open" set is one comprised of nothing but interior points.)

*While you do not need to draw up diagrams as did I, would you consider giving examples to support your definitions?

 A: The interval $(x-\epsilon,x+\epsilon)$ is also called the $\epsilon$-neighborhood of $x$. If we let $x=1$ (or alternatively $x=0$) then the intervals are of the form $(1-\epsilon,1+\epsilon)$ (or alternatively $(0-\epsilon,0+\epsilon)$), and we don't have that the interval is contained in $S$ (because it contains points that are $>1$ (or alternatively $<0$)). The rest of the points in $[0,1]$ are in fact interior points.
The flaw in your reasoning is the claim that if you get close enough to $1$ or $0$ you can get a situation where the $\epsilon$-neighborhood has finitely many points. This is not actually possible, not even for the rationals in $S$. Consider the $\epsilon$-neighborhoods of $0$, For any $\epsilon>0$ we have many point between $0$ and $\epsilon$, some examples include $\epsilon/2$, $\epsilon/3$ etc. as well as things like $\epsilon/\sqrt{2}$ and still many many more. In fact, there are infinitely many such points, no matter the $\epsilon>0$ you choose.
The definition of interior point is the definition of interior point, either you understand it or you don't. If the latter is the case, then looking at examples may help (feel free to ask away on this site too).
I can however, offer you a more abstract picture of what is going on. The notions of interior point and open set are fundamentally topological notions, meaning that they are determined by the topology of the space in question. In the case at hand, you are dealing with the standard topology on the reals. A topology for a set is literally the collect of sets defined to be open. In order to actually be a topology however, it must satisfy 3 criteria:
1: It must contain the empty set and the entire set.
2: Any arbitrary union of sets in the topology (this includes unions of infinitely many open sets as well as finitely many) must be in the topology (this can be phrased as: it is closed under arbitrary unions).
3: Any finite intersection of sets in the topology must be in the topology (closed under finite intersections).
Why this notion of a topology is important is not easy to explain to someone who hasn't dealt with sequences, series, and convergence rigorously (most introductory calculus classes are not rigorous in their treatment of this subject matter).
Topologies can be hard to describe because the collect of sets that are open is usually infinite, one method used to describe topologies is to use a basis. A basis for a topology is a subset of the topology (so a collection of some of the open sets), from which one can reconstruct all open sets via unions only.
The standard topology for the reals has as basis the collection of all open intervals $B=\{(a,b):a<b;a,b\in\mathbb{R}\}$.
So what about interior points? Well, the definition you are trying to understand can be phrased in terms of open sets instead of $\epsilon$-neighborhoods, after all, the latter are merely the basis elements for the standard topology of the reals. Hence one finds that a point in a set, $S$ is an interior point if there is an open set containing it that is entirely within $S$. It turns out that one can prove from these definitions that a set is open if and only if every point in that set is an interior point of that set. This gives a characterization of open sets in terms of interior points. One can also reverse this idea and obtain the notion of the interior of a (sub)set (of a topological space). The interior of a set $S$ is precisely the set of interior points of $S$, or equivalently, it is the union of all open sets entirely contained in $S$.
A: Here's an alternative way of thinking of it, although it is not rigorous and is informal.
Take two sets, $S \subset \mathbb R$ and $S^c =\{s \in \mathbb R| s \not \in S\}$.  $S$ can really be any set and $S^c$ or all the real  numbers that are not in $S$.  Every number/point in the "universe"/our system/the real numbers is either in $S$ or $S^c$ and no point is in both.
For example: Let $S = (0,1]U\{2\}$ and $S^c = =(-\infty, 0] \cup (1,2)\cup (2,\infty)$.
Now take any number/point $x$ and the set $S$.  Now one of 4 things can happen.
1) $x$ can be "utterly outside" the set $S$.  Which would be the same as saying it is "entirely inside" $S^c$.  
This would mean that we can always find some $\epsilon > 0$ so that $(x-\epsilon, x + \epsilon) \subset S^c$.  $-0.0000000001$ is such a number as $(-0.00000000009, -0.00000000011)\subset S^c$.  But $0$ is not.  Because $(0-\epsilon, 0+\epsilon) = (0-\epsilon,0]\cup (0, 0 + \epsilon)$ and $(0,0+\epsilon) \not \subset S^c$.
2) $x$ can be outside of $S$ but "on the edge of" $S$.  That means for every $\epsilon > 0$ no matter how small $(x-\epsilon, x+\epsilon)$ will contain some points in $S$.  $0$ is the only point like this in this set.  For any $\epsilon >0$  then the $(-\epsilon, \epsilon)$ will contain points in $(0, \epsilon)$ that will contain points in $S$.
3) $x$ can be inside of $S$ but not "entirely" inside $S$.  That means for every $\epsilon > 0$ no matter how small $(x-\epsilon, x + \epsilon)$ will contain so points in $S^c$.  $1,2$ are such points.
4) $x$ can be "entirely inside" of $S$.  i.e. $x$ is an "interior point".  That means there is always some $\epsilon$, maybe very small, so that $(x-\epsilon,x+\epsilon) \subset S$.  In our example, all the points $(0,1)$ are interior points.
Notice:  $(0,1)$ are the interior points of $S$.  $(-\infty, 1)\cup (1,2) \cup(2,\infty)$ are the interior points to $S^c$.  $0,1,2$ are not interior points to either.
A set is called "open" if all its points (if any) are interior points.  Neither $S$ nor $S^c$ is open.
Let $T = (0,1)$.  For every point $x \in (0,1)$ we know $0 < x < 1$.  Let $\epsilon = \min(\frac x2, \frac {1-x}2)$.  Then $0 < x-\epsilon \le \frac x2 < x < x + \epsilon \le x + \frac {1-x}2 < 1$.  So $x$ is an interior point.  So $x$ is open.
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By the way, a set would be called "closed" if all of it's "edge" points are in the set.
The "edge" points of $(0,1)$ are $0,1$.  $0,1$ are not in $(0,1)$ so $(0,1)$ is not closed.
The "edge" points of $[0,1]$ are $0, 1$.  $0,1$ are in $[0,1]$ so $[0,1]$ is closed.
Note $(0,1]$ is neither open nor closed.
Also note: if we let $S = \mathbb R$ then there are no "edge" points.  So $S$ is closed.  All points are interior points, so $S$ is open.  $S$ is open AND closed.
Also note:  If $S$ is open, then $S^c$ is closed.  If $S$ is closed, then $S^c$ is open.
This is because if $x$ is an interior point of $S$ it can not be an "edge" point of $S^c$.  So if $S$ is open, all the "edge" points of $S^c$ are not in $S$ so they are (if they exist) in $S^c$ so $S^c$ is closed.
Likewise if $x$ is in $S^c$ and it isn't an "edge" point of $S$ then it is "entirely inside" $S^c$.  If $S$ is closed then all the "edge" points of $S$ (if any) are in $S$ and none are in $S^c$.  So the only points left is $S^c$ (if any) are interior points.  So $S^c$ is open.
