Let us suppose the appropriate sequence exists and let us enumerate the fields of the chessboard as follows:

| 1 | 2 | 3 | 4 | 1 | 2 |
|---|---|---|---|---|---|
| 2 | 3 | 4 | 1 | 2 | 3 |
| 3 | 4 | 1 | 2 | 3 | 4 |
| 4 | 1 | 2 | 3 | 4 | 1 |
| 1 | 2 | 3 | 4 | 1 | 2 |
| 2 | 3 | 4 | 1 | 2 | 3 |

The length one moves go from odd to even number and vice versa. The length two moves go from even to a different even number or from odd to a different odd number. If we denote $P_1, P_2, \dots, P_{36}$ the numbers of visited fields, then it follows that among $P_2, P_3, P_4, P_5$ is each number (from $1$ to $4$) exactly once ($P_2$ and $P_3$ are different numbers with the same parity, and $P_4, P_5$ as well, only the parity is different). From the same reasons, any of the four numbers is among $P_{4k+2}, P_{4k+3}, P_{4k+4}, P_{4k+5}$ for arbitrary $k \in \{0, 1, \dots, 7\}$. Between the numbers $P_2, P_3, \dots, P_{33}$ is thus any of the numbers $1$ to $4$ exactly eight times.

The number $4$ is on the chessboard just eight times, thus none of $P_1, P_{34}, P_{35}, P_{36}$ can be $4$. The numbers $P_{34}$ and $P_{35}$ have the same parity and are different (they are the length two move apart). The number $4$ is not among them, therefore both must be odd. Then $P_{36}$ has to be even and $P_1$ as well. Thus it has to be number $2$.

The initial field ($P_1$) thus has to be one of the coloured fields on the left chessboard. One can repeat the arguments for the numbering of the right chessboard (just a rotation of the left one). Since no field has number $2$ on both chessboards, we come to a contradiction. The initial field cannot be chosen.