Definition of limit point using infinite sequence

I'm guessing the definition using the open ball is the following:

Suppose $M$ is a metric space with $E \subseteq M$. Then $x \in M$ is a limit point of $E$ if and only if, for all $\delta > 0$, $B(x; \delta) \cap E \neq \emptyset.$

These definitions are indeed equivalent. If the above definition holds, then for any $n \in \Bbb{N}$, we can take $\delta = 1/n > 0$ and we see that $$B(x; 1/n) \cap E \neq \emptyset.$$ We can then choose $x_n \in B(x; 1/n) \cap E$ for each $n$. That is, $x_n$ in $E$, and $$0 \le d(x, x_n) < \frac{1}{n}.$$ Thus, for any $\varepsilon > 0$, we see that $$n > \frac{1}{\varepsilon} \implies d(x, x_n) < \frac{1}{n} < \varepsilon.$$ Hence, $x_n \to x$, in $M$.

On the other hand, suppose instead the sequential definition holds, and we have a sequence $x_n \in E$ that converges to $x$. Then, for any $\varepsilon > 0$, there exists some $N$ such that $$n > N \implies d(x, x_n) < \varepsilon.$$ If we choose one particular $n$, we get $x_n \in E \cap B(x; \varepsilon)$, hence $E \cap B(x; \varepsilon) \neq \emptyset$.

Suppose, $x$ is a limit point of the sequence then for all $\varepsilon$ we have $D_{\varepsilon}(x)\cap E \neq \emptyset$. Fix a $r$ with $D_{r}(x)\cap E \neq \emptyset$ and consider the sequence $r>r_{i}>r_{i+1}>0$ for all $i$ such that $D_{r_{i+1}}(x) \subset D_{r_{i}}(x)$ and define $$A_{i}=D_{r_{i}}(x)-D_{r_{i+1}}(x)\neq\emptyset$$ in such a way that $r_{i}$ converges to $0$ such sequence $r_{i}$ exist as $\varepsilon$ is arbitrary clearly $D_{r_{i}}(x)\cap E \neq \emptyset$ for all $i$(from the defination of limit point only). Now choose a sequence $x_{i} \in A_{i}\cap E\neq\emptyset$(which can be garunted by choosing sequence $r_{i}$ in some optimal way). Hence we can construct a sequence $x_{i}$ that converge to $x$.

Algorithim to choose sequence $r_{n}$

$i)$ If $A_{i}\cap E\neq\emptyset$ choose $r_{i}$ and $r_{i+1}$ in a usual way.

$ii)$ $A_{i}\cap E=\emptyset$ then replace $r_{i+1}$ with $r_{i+2}$ and ckeck the above intersection is empty or not, if nonempty choose $r_{i+1}$ to be $r_{i+2}$, but if empty then replace $r_{i+2}$ with $r_{i+3}$ and repeat the process until some intersection is nonempty and suppose this process terminates at $r_{i+j}$ then choose $r_{i+1}$ with your $r_{i+j}$

$iii)$ After $r_{i+1}$ is choosen go for choosing $r_{i+2}$ in the same way as (ii) only difference is $A_{i+1}\cap E=\emptyset$, so keep procced until you choose $r_{i+2}$ with $A_{i+1}\cap E\neq\emptyset$,

$iv)$Thus select all of $r_{n}$ members.

The idea I am trying to say is, If $E$ is the union of regions(a region might contain only 1 point). But around $x$ there are infinitely many points in $E$. Construct ball around $x$ of different radius such that it forms infinitely many concentric circles(or balls or spheres) . By optimally picking the radius we can pick a point that is in $E$ and also in the region of difference of two distinct radius. The idea is to picking those radius such that every region of difference of two consecutive radius has a point that also in $E$