First sort the intervals in descending order by their lower bound, breaking ties in increasing order of their upper bounds. Let $dp(I)$ be the length of the longest sequence starting with interval $I$.

We can see that $dp(I)=max\{1+dp(J)$, for each interval $J$ completely contained in interval $I\}$. If $I$ does not contain any other intervals, then $dp(I)=1$.

Because of how we sorted the intervals, interval $I$ cannot contain interval $J$ if $I<J$. A direct implementation of the above idea has time complexity $\mathcal{O}(N^2)$ and does not get all points.

To efficiently calculate $dp(I)$, we introduce a new array of integers $A$ containing $1000000$ elements and initialized to zero. The value $A(hi)$ tells us the length of the longest sequence starting with an interval whose upper bound is exactly $hi$.

Now the relation $dp(I=[lo,hi])$ can be expressed as $dp(I)=max\{1+A(x),\text{for }lo\le x\le hi\}$. In other words, we find the length of the longest sequence whose upper bound is inside $I$. Again, the lower bound of that interval will also be within $I$, because the intervals are processed in descending order of lower bounds.

Finally, to efficiently find the largest element in the subsequence of $A$ with indices $[lo,hi]$, we can use an interval tree or Fenwick tree.