My intuition Cauchy (or convergent) sequences is that the smaller we make $\epsilon$ the further down the sequence we (might?) need to go to make that true. However, in the definition of Cauchy this is not a requirement and I am trying to understand why its not (and fix this intuition if its wrong and adjust it to the right in intuition since this seems an important conceptual detail about Cauchy sequences).
Recall the definition of a Cauchy sequence $(p_n)^{\infty}_{n=1}$:
$$ \forall \epsilon >0, \exists N_{\epsilon} : n_{\epsilon},m_{\epsilon} \geq N_{\epsilon} \implies d( p_{ n_{\epsilon} },p_{ m_{\epsilon}} ) < \epsilon$$
in plain English it says that for every accuracy epsilon we require, after some point in the sequence $N_{\epsilon}$ every pair of points are bounded within some (open) ball of radius $\epsilon$.
Consider $0<\epsilon_2<\epsilon_1$ (decreasing epsilon). To my understanding there are only two cases to consider:
1) When $N_{\epsilon_1}$ is already so far down the sequence that $\epsilon_1$ is actually to large so $d(p,q) < \epsilon_2 < \epsilon_1$
2) Or that $N_{\epsilon_1}$ is of non-trivial size and there are actually some elements $n_{\epsilon_1},m_{\epsilon_1} \geq N_{\epsilon_1}$ between $0<\epsilon_2< d( p_{ n_{\epsilon_1} },p_{ m_{\epsilon_1}} ) < \epsilon_1$.
In case 1) $N_{\epsilon_1}$ can be set to $N_{\epsilon_2}$ because for $n,m \geq N_{\epsilon_1} = N_{\epsilon_2}$ we have $d(p,q) < \epsilon_2 < \epsilon_1$ which is what we required (if we know $N_{\epsilon_1}$ of course). So we do not need to increase $N$ in this case.
In case 2) we first note that its impossible for all the terms $n_,m \geq N_{\epsilon_1}$ to be between $0<\epsilon_2< d( p_{ n_{\epsilon_1} },p_{ m_{\epsilon_1}} ) < \epsilon_1$. If this were the case then the sequence wouldn't be Cauchy because there would be some epsilon (namely $\epsilon_2$) such that its impossible to find an $N$ s.t. they are all less than $\epsilon_2$. Namely they all get stuck at a distance at least $\epsilon_2$ but we require to be able to always find an $N$ for which their distance is at most some positive constant (which we showed its impossible for $N_{\epsilon}$). Ok if this is not the case then we call $E$ the pair of elements of the sequence such stuck in that interval $(\epsilon_2,\epsilon_1)$ i.e.
$$ E = \{ (p,q) \in (p_n)^{\infty}_{n=N_{\epsilon_1}} \mid \epsilon_2 < d(p,q) < \epsilon_1 \}$$
this set must be finite. If it weren't then it would have all the elements of the sequence and again it would imply the sequence is not Cauchy again. Now consider going through all the elements of $E$ and collecting their indices in the set $I$. $I$ must also be finite. Thus it must have a maximum $M = \max I$. Now let:
$$ N^* = M+1 = \max I + 1$$
note $N^*$ is defined to be the index of an element of the sequence such that when compared to any other element its not in the interval $(\epsilon_2,\epsilon_1)$. This element (and its index) must exist. If it didn't then $d(p_{N^*},q) $ would be in $(\epsilon_2,\epsilon_1)$. This can't be true because if it were in $I$ then $N^*$ would be larger than $M$ but $M$ wouldn't be the max (but its defined to be the maximum index from $E$). Therefore, $N^*$ must not be in $I$ and thus we can set $N^* = N_{\epsilon_2}$. Which shows that if $\epsilon_2$ is smaller than $\epsilon_1$ then unless $\epsilon_1$ is too larger, we will need to go down the sequence to get a new more meaningful $N$ if the sequence is Cauchy.
If possible I'd love to know if my proof is correct of course.