Proof 1: (Formal)
Case 1: If there is any group of $6$ or more mathematicians who are asleep at same time at some point during the lecture, the problem is solved.
Case 2: Assume there is no group of $6$ or more mathematicians who are asleep at same time at some point during the lecture. This implies that:
(a) There are no more than 5 mathematicians asleep at same time at some point during the lecture.
(b) Denote each mathematician with an integer number, such as $M_i$, where $1\le i\le 26$, based on the order of starting time of their sleeping periods. For example mathematician $M_1$ is the first one who goes asleep, and mathematician $M_{26}$ is the last one who goes asleep.
(c) Additionally, denote the sleeping period of mathematician $M_i$ as an interval $(a_i, b_i)$, where $a_i$ is the starting time, $b_i$ the ending time, and $a_i<b_i$ ($i=1,2,\ldots,26$). By (b), we have $$a_1\le a_2 \le a_3 \le \ldots \le a_{25} \le a_{26}$$
Therefore, for any 6 consecutive starting times of sleeping periods of 6 mathematicians, $M_i, M_{i+1}, M_{i+2}, M_{i+3}, M_{i+4}, M_{i+6}$, their sleep periods will be $${(a_i, b_i), (a_{i+1}, b_{i+1}), (a_{i+2}, b_{i+2}), (a_{i+3}, b_{i+3}), (a_{i+4}, b_{i+4}), (a_{i+5},b_{i+5})}$$
There exists at least one $(a_j, b_j)$, $i\le j\le i+4$, so that it does not overlap with $(a_{i+5}, b_{i+5})$, i.e. $a_j<b_j<a_{i+5}$; otherwise these $6$ sleeping periods all overlap with each other, in contradiction to the assumption (a). This also implies that $(a_j, b_j)$ does not overlap with any sleeping period $(a_k, b_k)$ where $k\ge i+5$, as $a_j<b_j<a_k<b_k$, which translates to:
(d) The sleeping period of mathematician $M_j$ does not overlap with the sleeping period of any mathematician $M_k$, where $k\ge i+5$.
(e) In this way, we can find $5$ mathematicians, one each from $5$ distinct groups of mathematicians ${M_i, …, M_{i+4}}$ where $i=1,6,11,16,21$, so that the sleeping period of these $5$ mathematicians do not overlap with each other.
By (d), the sleeping period of mathematician $M_{26}$ does not overlap any of the $5$ mathematicians selected in (e). Therefore, we have found $6$ mathematicians whose sleeping periods do not overlap with each other.
Combining the above two cases completes the proof.
Proof 2: (Informal)
Assume that there are $5$ private rooms, in each room only one person can sleep; and there is a public room that can sleep $21$ people.
Whenever a mathematician falls asleep, they are instantly teleported to a private room if available, or to the public room if all private rooms are occupied.
Whenever a mathematician wakes up, they are instantly teleported back to the original place.
Case 1: If there is a mathematician at any point sleeping in the public room, it means that all private rooms were occupied, so $5$ other mathematicians were asleep in the private rooms. Therefore, there are at least $6$ mathematicians fell asleep at the same time.
Case 2: If there is no mathematician at any point sleeping in the public room, it means that every mathematician was asleep in $1$ of $5$ private rooms at sometime. By pigeonhole principle, at least $6$ mathematicians were sleeping in the same private room, as $5\times 5=25<26$ and no two of these $6$ mathematicians could have been asleep at the same time in that private room.
Combining the above two cases completes the proof.
mathematicians falls asleep exactly once, and stays asleep for a nonzero amount of time. Each mathematician is awake at the moment the lecture starts, and the moment the lecture finishes. Prove that there are either 
mathematicians such that no two are asleep at the same time, or