Understanding the definition of a limit point.
Let $\mathbb{R}$ be given the usual metric.
The idea of limit points of a subset $A \subset \mathbb{R}$ is to say that in the neighbourhood of such points, I can find at least one other point different from that point which is also in $A$.
So if $x \notin A$, then $x$ is a limit point of $A$ iff in any $\varepsilon$-neighbourhood of $x$, I can find at least one point of $A$. This is precisely your modified definition. So yes, it should be valid.
Addendum:
For the edited part: Note the condition $a_n \neq x$ for all $n \in \mathbb{N}$ carefully. This condition precisely prevents the constant sequence that you mentioned.
If this condition is not present, then every point in a set $A$ could be a limit point of $A$. This is not necessarily true. Consider the following set \begin{equation} S = \{ 1/n : n \in \mathbb{N}\}. \end{equation} Then none of the points in $S$ is a limit point of $S$. Moreover, the only limit point of $S$ is $0$ which is not in $S$.
Finally, the definition does not fail with the constant sequence as it is sufficient to find just one such sequence that satisfies $x = \lim a_n$.
"I am trying to understand what "other than x" " What else could it mean? If $x \in A$ but it is a singleton point and there is a neighborhood around $x$ that contains no other point of $A$ except $x$ then that is not a limit point. Even though every neighborhood does contain $x$ itself. And $x \in A$.
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Consider the the set $\mathbb Z\subset \mathbb R$ and the "usual" metric.
Is $1.999$ a limit point of $\mathbb Z$?
If you take any $\epsilon$ no matter how small, then the $\epsilon$-neighborhood around $2$ is $(1.999-\epsilon, 1.999+\epsilon)$. And if $\epsilon \le 0.001$ then $there is no integer in that neighborhood.
So $1.999$ is not a limit point.
Now does $\mathbb Z$ have any limit points?
Before we answer that, let's ask, SHOULD $\mathbb Z$ have any limit points?
$\mathbb Z$ is a collect of distinct points separated from each other. It "shouldn't" have any limit points points, because a limit point should be a point the no matter how close you get to it, there's going to be a point in the set right there close to it. And with $\mathbb Z$ if we start taking epsilons less than $1$ then all the points in t set are all going to be isolated and alone. If you start getting closer than one to any point, that points should be isolate and apart from all integers.
So what about $2$? Is $2$ a limit point of $\mathbb Z$. Well, if you take any $\epsilon: 0 < \epsilon < 1$ then the $\epsilon$-neighborhood $(2-\epsilon, 2 + \epsilon)$ doesn't have any integers ....
... except $2$.... Evey $\epsilon$-neighborhood of $2$ has $2$ itelf in it.
So be your definition, $2$ is a limit point.
But.... it shouldn't be. To say "there is always an integer close to $2$" shouldn't count for $2$ itself. It's.... weird ... to say that $2$ is "close to" $2$. This misses the point of limit points entirely.
A limit point should be that every neighborhood of $x$, no matter how small, you can find a element of $A$. But $x$ itself doesn't count. You need to find an element of $A$ that is "close to" $x$; not the point $x$ itself.
If $x \not \in A$ that's not an issue. Everypoint of $A$ that is close to $x$ won't be equal to $x$ because $x$ isn't in $A$. But if $x\in A$ then to be a limit point the every $\epsilon$-neighborhood of $x$ will contain a point of $A$ (other that $x$ itself).
That's really all it means.
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2). That alternative definition fails if $\lim a_n = x$.
Consider $\mathbb Z$ which has no limit points. ANd $x = 1.999$ and $y=2$.
Let $a_i \in \mathbb Z$. And let $\lim a_i = 1.999$. That is clearly impossible. SO there is no sequence of integers hows limit is $1.999$, so $1.999$ is not a limit point. That makes sense. You can't "get close" to $1.999$ because you will always be at least $0.001$ away.
Now le $a_i \in \mathbb Z$ and $a_i\ne 2$. Then $\lim a_n = 2$ is imposibble. You can't "get close" to $2$ because you will always be $1$ away. So $2$ can't be a limit point.
But what if $a_1 = 1, a_2 =3$ and for $i > 2; a_i =2$.. Then $\lim a_n = 2$. But that doesn't count. You aren't "getting close to $2$". You are jumping across a brook and landing smack dab on $2$. That's not "getting close". That's stomping around with you big boots on and being there.