OF COURSE HE IS RIGHT, HE IS A PROFESSOR.
We will prove the problem for 8 friends instead of 5 and see that this is the optimal bound.
Proof.
Let us imagine that each of the professor's friends has a twin who visits him on the days when the sibling does not. Then we have $40$ people in total and every day exactly $20$ of them visit the professor. Now if we prove that there exist days $D_1, D_2$ and $8$ people who have visited the professor on both days $D_1$ and $D_2$, then the problem will be solved. Indeed, it is impossible that we choose both twins of a pair in these $8$ people, because at most one of them can visit the professor on both days. If some of these $8$ people are not the professor's friends, just exchange them with their twins and we are done.
Next, let us label the friends and their twins with #1, #2, ... , #40 such that #j is the twin of #(j+20).
For each $i\leq 6$, denote with $x_i$ the $40$-dimensional vector
$$x_i=(\delta_{i,1}, \delta_{i,2}, ... \delta_{i,40}),$$
where $\delta_{i,j}$ is $1$ if person #j visited the professor on day i and $0$ otherwise. If we assume that no $8$ people visited the professor on the same 2 days, we have $\langle x_k, x_l\rangle \leq 7$ for all pairs $(k,l)$, where $\langle \cdot, \cdot \rangle$ is the regular scalar product.
Now we have
$$\langle V, V\rangle=\langle \sum_{i\leq 6} x_i, \sum_{i\leq 6} x_i \rangle = \sum_{i \leq 6} \langle x_i,x_i\rangle + \sum_{i\neq j} \langle x_i, x_j \rangle \leq 6.20+30.7=330.$$
Let $V=(V_1, V_2,...,V_{40})$. For every $j\leq 20$ we have $V_j + V_{j+20}=6$ and therefore $V_j^2+V_{j+20}^2\geq 18$. Therefore
$$\langle V, V \rangle = \sum_{j\leq 40}V_j^2 \geq 20.18=360,$$
which gives us a contradiction.
In order to see that $8$ friends is optimal, just consider the following example courtesy of @The Dark Truth (may want to upvote him as well for this):
1111 1111 1100 0000 0000
1111 0000 0011 1111 0000
1000 1110 0011 1000 1110
0100 1001 1010 0110 1101
0010 0101 0101 0101 1011
0001 0010 1100 1011 0111