Let $(X,\|\cdot\|)$ denote the space you are describing.
For each $k\in\mathbb{N}$ define $Y_k$ to be the subspace of $X$, consisting of sequences $x=(x_j)$ such that $x_j=0$ for all $j>k$. It is easy to see each $Y_k$ is closed in $X$ and $X=\cup Y_k$.
However, $Y_k$ has empty interior. To see this, take $x=(x_j)\in Y_k$ and $\epsilon>0$. Then we can always define $y=(y_j)$ such that $y_j=x_j$ for $j\le k$, $y_{k+1}=\epsilon$ and $y_j=0$ for other $j$. Then $\|y-x\|<\epsilon$ but $y$ is not in $Y_k$.
Now each $Y_k$ is nowhere dense, so their countable union $X$ cannot be complete as a result of Baire's theorem.
Or, if you prefer concrete examples. Take $x^n$ to be the sequence whose first $n$ terms are $(1,\frac{1}{2},\dots,\frac{1}{n})$ and all other terms vanishes. You can see this $x^n$ is Cauchy in $X$, but you should feel they converge to the sequence $(1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{n},\frac{1}{n+1},\dots)$, which is not in $X$.
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 $d$.
Best Answer
Consider the sequence $x_n=-n$. Clearly it doesn't converge. For, suppose $x_n \to x$ for some real number $x$. Then $|e^{-n}−e^x|→0$. Hence, $e^x=0$, which cannot be.
However, $N<n<m \implies |e^{x_n}-e^{x_m}|=e^{-n}-e^{-m}<e^{-N} \to 0$ as $N \to \infty$. So the sequence is Cauchy.