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    $\begingroup$ GeometricFunctions`BinarySearch[S, #] & /@ list1 almost does what you want. $\endgroup$ Commented Sep 1, 2017 at 16:38
  • $\begingroup$ Sidenote: Does BinarySearch really use binary search? Because the running time seems to increase linearly with the length of the list?? $\endgroup$ Commented Sep 1, 2017 at 18:28
  • $\begingroup$ Could you please clarify if you mean "first $s \in S$ with $s > l$" or "last $s \in S$ with $s <l$"? $\endgroup$ Commented Sep 1, 2017 at 20:23
  • $\begingroup$ i don't get why the first number for 6.0669 is at position 2 and not 1, do you mean the higher element of s such that s<l ? $\endgroup$ Commented Sep 2, 2017 at 3:50
  • $\begingroup$ Yes, I had a mistake in the description, hopefully it's clear now. $\endgroup$ Commented Sep 2, 2017 at 4:12