This is how I understand it.
The definition of convergent sequence in a normed space is that:
for every $e>0$ there is $d>0$, such that $|a_n-a|<e$, when $n>d$.
This, translated to neighborhoods, means that
for every neighborhood of $a$ there is a tail of the sequence completely inside the neighborhood.
For the normed case we can produce a system of neighborhoods of a point that is countable, totally ordered by inclusion, and such that for every neighborhood of the point there is one in the system that is inside. Then one just needs that for each element of the system of neighborhood to contain a tail of the sequence.
So, in the absence of a norm we just need to imitate this. We consider a neighborhood system, which need not be countable. This forces us to consider $a_I$ for arbitrary sets $I$. Next the system may be forced to not be totally ordered by inclusion. So we need to define what would be the tail of the 'sequence' $a_I$. We define it by the properties that we need them to satisfy.
If $U_1$ and $U_2$ are neighborhoods of the system of neighborhoods, we want tails of $a_I$ to be in $U_1$ and in $U_2$. But $U_1\cap U_2$ is also open and therefore must also contain a tail of $a_I$.
So, for $U_i$, $i=1,2$, we have $d_1\in I$ such that for $j>d_i$ $a_j\in U_i$.
We also want something like this for $U_1\cap U_2$, but we would like it to be similar as the usual definition of limit.
We want that for every neighborhood of $a$ "eventually" all elements of the sequence are in there. So, we need to impose, at least, that the tails $j\geq d_i$, $i=1,2$ merge. Otherwise the tails for $U_1,U_2$, and $U_1\cap U_2$ may not have anythign to do with eachother. Imposing that these tails have non-empty intersection means that we have $d$ such that $d\geq d_1$ and $d\geq d_2$. So, we impose this property on $I$.
For every $d_1,d_2\in I$ there is $d\in I$ with $d\geq d_1$, and $d\geq d_2$.