Let $c$ be the space of convergent sequences over $\Bbb R$ with the $\sup$ norm.
Let $X$ be the subset of constant sequences $s=2^m3^n,2^m3^n,\ldots:m,n\in\Bbb Z$, for example $s=1,1,1,1,\ldots$ with the same metric $d$ (possibly including the sequence $0,0,\ldots$).
Is $X,d$ a complete metric space?
I don't really know how to tackle this. My first thought is that since the sequences are constant, the space is no different to using the absolute value metric on singletons $2^m3^n\in\Bbb Q^+$. My second thought is that the set $2^m3^n$ is dense in $\Bbb R$ so this is unlikely to be a discrete space so no trivial answer is clear to me.
I tried engineering Cauchy sequences and I can easily make ones that converge to $0,0,0,\ldots$ so I think the constantly zero sequence is needed for it to be complete. I can't really make Cauchy sequences that converge to anything other than $0$ and some element of $X$, but I can't show it's possible or impossible to create others either.
Let $A=\{2^m3^m\mid n,m\in\Bbb Z\}\subset \Bbb R$ and consider the map $\varphi: \Bbb R\to c$, $x\mapsto (x,x,x,x,...)$. It is easy to check that the $d(\varphi(x),\varphi(y))=|x-y|$, ie the map is an isometry onto its image. As such $\varphi(A)$ will be complete if and only if $A$ is complete.
That $A$ is complete, is not true. There are likely many reasons why $A$ is not complete, for example one can see that there is a sequence $x_n\in A$ with $x_n\to 5$ but $5\notin A$. The statement $5\notin A$ is elementary, while the construction of such a sequence follows by noting that $\{\lfloor m\log_2(3)\rfloor \mid m\in \Bbb Z\}$ is dense in $[0,1]$ (since $\log_2(3)$ is irrational) and as such for any $\epsilon$ you can find $n,m$ so that $|n- m\log_2(3)| ≤ \log_2(5) + \epsilon$, implying that $5\cdot 2^{-\epsilon}≤\frac{2^n}{3^m} ≤ 5\cdot 2^\epsilon$, giving that you can approximate the point $5$ arbitrarily well.