Negative exponential distance Let $X := \left\{(a_k)_{k \in \mathbb N}, a_k \in \mathbb C\right\}$. Let $d\left( (a_k)_{k \in \mathbb N}, (b_k)_{k \in \mathbb N} \right) := e^{-u}$ with $u$ the smallest integer $k$ such that $a_k \ne b_k$ be a distance on $X$. Is $(X, d)$ compact/complete/connected?
Here's my not very rigorous take : $X$ is not complete since it's not even a normed space (I don't think there exists a norm that induces $d$). It is not compact since $\left((1, \dots ), (2, \dots), (3, \dots), \dots\right)$ is a sequence of elements of $X$ that does not have a convergent subsequence. I can't find an argument for connectedness, any ideas?

You are correct that $X$ is not compact, by exactly the example you mention.
On the other hand, $X$ is in fact complete.  Observe that if $d((a_k), (b_k)) < e^{-n}$, then $a_i = b_i$ for all $i \leq n$.  It follows that if $(\mathbf{a}_n)_{n\in \mathbb{N}} = ((a_{n,k})_{k\in\mathbb{N}})_{n\in \mathbb{N}}$ is a Cauchy sequence, then for each $k$, the sequence $(a_{n,k})_{n\in\mathbb{N}}$ is eventually constant (call its eventual value $b_k$).  One can then show that the sequence $(b_k)_{k\in \mathbb{N}}$ is the limit of $(\mathbf{a}_n)_{n\in \mathbb{N}}$.
Finally, $X$ is not connected.  Indeed, for any $\mathbf{a} = (a_k)_{k\in \mathbb{N}}$, the $\varepsilon$-ball $B$ centered at $\mathbf{a}$ is disjoint from the union of the $\varepsilon$-balls centered at at every $\mathbf{b} \not \in B$.  These are open, so it follows that $X$ is not connected.
