Understanding the idea of a Limit Point (Topology)
The problem seems to be that you’ve got your quantifiers backwards. It isn’t that there’s a single $q\in E$ such that $d(p,q)<r$ for arbitrarily small $r$. Rather, for each positive $r$, no matter how small, there is a $q\in E$ such that $d(p,q)<r$. The choice of $q$ depends on $r$. In symbols, this is the difference between
$$\exists q\in E\setminus\{p\}\;\forall r>0 (d(p,q)<r$$ and $$\forall r>0\;\exists q\in E\setminus\{p\}(d(p,q)<r\;.$$
For a simple example, let $$E=\left\{\frac1n:n\in\mathbb{Z}^+\right\}\;,$$ and let $p=0$. For $r=0.1$, you can take for $q$ any $1/n$ with $n>10$. With $r=0.01$, on the other hand, you’ll need to choose a $1/n$ with $n>100$. And so on.
I’d also say that your picture is inside-out: you should think of circles of decreasing radius squeezing in closer and closer to $p$. Then $p$ is a limit point of $E$ if within each of those circles, no matter how close to $p$, there is at least one point of $E$ different from $p$ itself.
Added:
Now let’s take a look at your three ‘general operations’.
$$d(r_{n},r_{n-1}) \to 0$$
If you choose a sequence $\langle r_n:n\in\mathbb{N}\rangle$ converging to $0$, then it will automatically be the case that $d(r_{n},r_{n-1}) \to 0$ as $n\to\infty$, but this is a side-effect, not something on which you should focus. What’s important is that $r_n\to 0$ as $n\to\infty$; if that’s the case, and if for each $n\in\mathbb{N}$ you have a $q_n\in E$ such that $q_n\ne p$ and $d(p,q_n)<r_n$, then $p$ is a limit point of $E$.
$$\forall r \in \mathbb{R} \exists q \in E \big(d(p,q_{n})<r_{n}, n \in \mathbb{N}\big)$$
As written, this doesn’t make sense: how are the single $r$ and $q$ in the quantifiers related to the $r_n$ and $q_n$ in the quantified statement? You could correctly write any of the following, since all of them say that $p$ is a limit point of $E$:
$$\begin{align*} &\forall r>0 \exists q(r)\in E\big(q(r)\ne p\text{ and }d(p,q(r))<r\big)\\ &\forall n\in\mathbb{Z}^+\exists q_n\in E\left(q(r)\ne p\text{ and }d(p,q_n)<\frac1n\right)\\ &\forall n\in\mathbb{N}\exists q_n\in E\left(q(r)\ne p\text{ and }d(p,q_n) < \frac1{2^n}\right) \end{align*}$$
Indeed, if $\langle r_n:n\in\mathbb{N}\rangle$ is any sequence of positive real numbers converging to $0$, you could take
$$\forall n\in\mathbb{N}\exists q_n\in E\big(q(r)\ne p\text{ and }d(p,q_n) < r_n\big)$$
as your definition of ‘$p$ is a limit point of $E$’.
$$d(p,q_{1}) \to 0$$
As written this makes no sense, since $p$ and $q_1$ are single, fixed points: there is no sequence here. Did you mean $d(p,q_n)\to 0$? That isn’t enough as it stands, because it says nothing about the nature of the $q_n$. What does work is this:
A point $p$ is a limit point of a set $E$ if and only if there is a sequence $\langle q_n:n\in\mathbb{N}\rangle$ of points of $E\setminus\{p\}$ such that $d(p,q_n)\to 0$ as $n\to\infty$.
(This of course assumes that there is a metric $d$; this definition doesn’t work for topological spaces in general.)