I think it should be solved using the pigeonhole principle. The answer is:
A $21-$gon has $189$ diagonals. If through a point in the plane, we draw parallels to these diagonals, $2 × 189 = 378$ adjacent angles are formed. The angles sum up to $360^\circ$, and thus one of them must be less than $1^\circ$.
I do not understand what is meant by "we draw parallels to these diagonals" nor where $360$ came from. The only solution I could think of was drawing all the diagonals through the same point and then counting how many angles "make up" the full circle of angles; I do not know if this is the correct interpretation.

The reasoning given is correct. By "we draw parallels to these diagonals," it means "we draw the unique line parallel to a given diagonal, for every single diagonal."
However, if $0-$degree angles are not counted as "angles", it is false.
Proof: Every diagonal of a regular $21$-gon is in one of at most $21$ directions which have angles of $8\frac{4}{7}$ degrees with each other. Therefore, the minimum angle in this case is at least $8\frac{4}{7}$ degrees, which is much greater than $1$ degree.
In fact, $8\frac{4}{7}$ degrees is optimal, since all directions of the diagonals that connect a vertex and another vertex $10$ edges away are different (since they either intersect at a vertex or inside the polygon, and therefore cannot be parallel), and $8\frac{4}{7}$ degrees is optimal for $21$ different directions.