You are all right, the metric $d$ makes your metric space being discrete topological space. In the latter, convergence of $(x_n)$ means that there exists $a$ such that $x_n = a$ for all $n$ big enough. You perfectly showed this using $\epsilon = \frac12<1$. As a result, the sequence $\frac1n$ does not converge.
Any Cauchy sequence thus has to have only constant terms after some $n$ because of the Cauchy condition $|x_n -x_m|<\epsilon$ for all $m,n$ big enough. As a result, you obtain a desired sequence with only constants in the tail, so it clearly converges thanks to the first argument. And surely, the sequence $\frac1n$ is not Cauchy w.r.t. the metric $d$.
“Intrinsically convergent” wouldn’t mean literally convergent, because we have a term for that already, “convergent.” So, as a technical term you shouldn’t take “intrinsic” quite so literally.
Not every Cauchy sequence in the rationals converges in the rationals, but they do in the real numbers. So, while these sequences don’t converge in $\mathbb Q,$ there is a sense in which $\mathbb Q$ is “missing points” to cover these cases, and that they really do converge “some place.”
More generally, given a metric space, $(X,d),$ there is a unique space $(\overline X,d’),$ called the completion of $X,$ which contains $X$ and in which all Cauchy sequences converge. $\overline X$ is in some sense a “natural” extension of $X.$
The completion of the rational numbers is the reals.
For another example, if $$X=\{(x,y)\in \mathbb R^2\mid x^2+y^2<1\}$$ with the usual Euclidean distance, then the completion is identical to $\{(x,y)\mid x^2+y^2\leq 1\}.$
The Cauchy sequences in $X$ that don’t converge are instead intrinsically converging to the points on the boundary.
So, while a Cauchy sequence in $X$ does not converge, there is a single “natural” extension of $X$ in which all Cauchy do converge.
Essentially, a metric space where some Cauchy sequences do not converge can be thought of as “missing points” in the same way the rationals are missing the irrational reals, or $X$ is missing the boundary.
These points are “intrinsically” there, in that they exist in a natural and unique extension of the incomplete space.
If we wanted to borrow a word used in abstract algebra, we might use the term “Ideally Convergent.”
Best Answer
First of all notice that (according to your example) this metric space is not complete. So there are non-convergent Cauchy sequences. If $x_{n_k}\to\infty,$ then it is not convergent in $\Bbb{N}$
(It may convergent in $\Bbb{N}\cup\{\infty\}$). So Any convergent sequence $(x_n)$ must be bounded.
Let $N\in\Bbb{N}$ be fixed. Then for any sequence $(x_n)$converge to $N,$ we have $$d(x_n,N)=\left\lvert{\frac{1}{x_n}−\frac{1}{N}}\right\rvert\lt\epsilon$$ for sufficiently large $n\in\Bbb{N},$ where $\epsilon\gt 0$ is arbitrary.
In other words $$x_n\gt\dfrac{N}{\epsilon N+1}.$$ By letting $\epsilon\to 0$ we can conclude that $x_n\ge N$ for large $n.$
Again choosing $\epsilon =\dfrac{1}{N(N+1)}$ one can prove that $x_n\lt N+1$ for large $n$ values.