Proof Noting that $S$ is the longest sequence of numbers satisfying the condition. Hence, if we add a term, $0$ or $1$, after the last term of $S$, there will occur two identical sections of successive $5$ terms in $S$, and that is, there exist $i \neq j$ such that
$$(a_i, a_{i+1}, \dots, a_{i+4}) = (a_{n-3}, a_{n-2}, \dots, a_n, 0),$$
$$(a_j, a_{j+1}, \dots, a_{j+4}) = (a_{n-3}, a_{n-2}, \dots, a_n, 1).$$
If $(a_1, a_2, a_3, a_4) \neq (a_{n-3}, a_{n-2}, a_{n-1}, a_n)$, then
$$1 < i, j < n-3,\ i \neq j,$$
and
$$\begin{aligned} (a_i, a_{i+1}, a_{i+2}, a_{i+3}) &= (a_j, a_{j+1}, a_{j+2}, a_{j+3}) \\ &= (a_{n-3}, a_{n-2}, a_{n-1}, a_n). \end{aligned}$$
Now, consider $a_{i-1}$, $a_{j-1}$ and $a_{n-4}$, among which there must be two identical terms. This causes two sections with successive $5$ terms respectively in $S$ to be identical. It leads to a contradiction.
Therefore, the proposition holds.