I think I have an algorithm but I'm not sure if it's correct.
We want to maintain an augmented binary search tree $S$ on each interval end point. The splitting element will always be the median end point so that the total height of our tree in the end is in the order of $\log_{2}N$.
In every node we use some data structure to store all intervals that go through the splitting element.
Assuming nothing more for our data structure, we can see that if $x_q$ is the query point, every time we visit a node we check to see if $x_{q}$ is to the right or to the left of the splitting element. If it is to the left, then no interval stored in any node of the right subtree will intersect $x_{q}$ so we will only need to check the left sub tree. The same applies to the case where $x_{q}$ is to the right of the splitting element.
So far we can find all relevant intervals in $O(\log n)$ time, but to find the exact intervals that intersect $x_{q}$, we have to see what's going on in this structure that we have for every node, so we also have to query each such structure along a root to leaf path to find exactly the intervals that intersect $x_{q}$
To do this, we can have two sorted lists $L$, $R$ on every node, each list will store the same intervals but $L$ will store them in a sorted order according to the left point and $R$ according to the right point.
Then whenever we visit a node we see if $x_{q}$ is to the left of the splitting element. If that's the case, we do a binary search on $L$ to find the right most interval that contains this point, since we know the index of this interval in $L$ we can find the total amount of intervals.
We process similarly $R$ if $x_{q}$ is to the right of the splitting element.
This algorithm will give us a $O(\log ^2 N)$ complexity.
