Let $\mathcal{T}$ the set of sequences $t\mapsto at^\alpha$, with $t\in\mathbb{N}$, $\alpha>0$ and $a\in\mathbb{R}$. If necessary, it is possible to add boundedness assumptions, such that $\alpha\geq \alpha_{\min}>0$ and $0<a_{\min}\leq |a|\leq a_{\max}$.
I'm looking for a metric $d$ on $\mathcal{T}$ such that $$\forall\varepsilon>0,\, \exists \alpha(\varepsilon)>0,\, M(\varepsilon)>0,\, t(\varepsilon)\in\mathbb{N} : \left(d(u,v)>\varepsilon\right) \implies \left(\forall t\geq t(\varepsilon),\, |u(t)-v(t)|\geq M(\varepsilon)t^{\alpha(\varepsilon)}\right).$$
In other words, as soon as $d(u,v)>\varepsilon$ the rate of divergence of $u$ and $v$ is bounded from below by some function of $\varepsilon$ only.
I tried to use the metric $d(at^\alpha,bt^\beta) = \max(|a-b|,|\alpha- \beta)$, but it fails, as for example $d\left((1+2\varepsilon)t,t^{1+\frac{1}{n}}\right)>\varepsilon$, yet these two functions don't satisfy the desired property, as their rate of divergence can be made as small as possible by letting $n$ go to $\infty$.
So I'd like to either find a metric that works, or at least to prove that it does or does not exist.
I'll write your $t(\epsilon)$ as $t_0(\epsilon)$ to avoid ambiguity. The condition implies that if $u(t)=v(t)$ for some $u,v,t,$ then $t<t_0(\tfrac12 d(u,v)).$ In particular $d(nt,t^2)\to 0$ as $n\to\infty,$ because the two functions are equal at $t=n.$ (To clarify: I mean $d(u_n,v)\to 0$ where $u_n$ is the monomial $nt^1$ and $v$ is the monomial $t^2.$) This means $nt\to t^2,$ but by the same logic $nt\to t^3,$ which means $d(t^2,t^3)=0,$ so $d$ is not a metric.