If $(x_n)$ is a sequence, and $x$ is the limit, provided that it exists, in the definition, we have a part where $|x_n-x|<\epsilon$.
Now we also have a theorem:
If $a\in\Bbb{R}$ such that $0\le{a}<\epsilon$ for every $\epsilon>0$, then $a=0$.
Why can't we compare $a=|x_n-x|$ and therefore conclude that $|x_n-x|=0,\forall{n}\ge{K(\epsilon)}$ ?
In my textbook, the said theorem is used to prove the uniqueness of limit. So why won't the same property apply in the definition?
Edit: I corrected my condition of $n$, which I by mistake wrote $\forall{n}\in\Bbb{N}$.
In your theorem about $0 \leq a < \epsilon$ for all $\epsilon>0$, the constant $a$ does not change. That is, $a$ is a particular real number. In contrast, using $a=|x_n-x|$ does not make sense, since the left-hand-side is $a$ (which does not depend on the index $n$) while the right-hand-side is $|x_n-x|$ (which depends on the index $n$).
The definition of limit of $x_n$ being a real number $x$ says that for all $\epsilon>0$, the inequality $|x_n-x|<\epsilon$ is true for all sufficiently large values of the index $n$ (not necessarily for all values of $n$).
For example, if $x=0$ and $x_n=1/n$ for $n \in \{1, 2, 3,...\}$, then we have:
The inequality $|x_n-0|<1/1000$ is true for all $n \in \{1001, 1002, 1003, ...\}$.
The inequality $|x_n-0|<1/34000$ is true for all $n \in \{34001, 34002, 34003, \ldots\}$.
In general, the inequality $|x_n-0|<\epsilon$ is true for all integers $n$ that satisfy $n>1/\epsilon$.
So if $x_n=1/n$ for $n \in \{1, 2, 3, ...\}$, we have $\lim_{n\rightarrow\infty} x_n = 0$. However, certainly it is not true that $|x_n-0|=0$ for all $n \in \{1, 2, 3, ...\}$.