For a string $U$, let $|U|$ denote the length of that string. All substrings mentioned here are nonempty and of length $\le p$, and the given $W$ can be regarded as a circular string of length $p$, namely $x_0x_1x_2\dots x_{p-1}$; since such $W$ is finite, we can define the multiplicity of any substring $U = a_0a_1\dots a_{k-1}$ occurring in $W$ as
$$m(U) = |\{i|0 \le i \le p-1 \text{ and } a_0a_1\cdots a_{k-1} = x_i x_{i+1} \cdots x_{i+k-1}\}|$$
where all subscripts are taken (mod $p$). In other words, the string $U$ occurs in $W$ if and only if $m(U) \ge 1$. Moreover we have the equalities
$$m(U) = m(aU) + m(bU) = m(Ua) + m(Ub),$$
for any string $U$ of length $\le p-1$. So when $|U| \le p-1$, we only need to show: $m(aU), m(bU), m(Ua), m(Ub) \ge 1$. For $|U|=p$ we treat it specially: if some $U = a_0a_1\cdots a_{p-1}$ has $m(U) \ge 2$, then there exist $0 \le i < j \le p-1$ such that
$$a_0a_1\cdots a_{p-1} = x_i x_{i+1} \cdots x_{i+k-1} = x_j x_{j+1} \cdots x_{j+k-1};$$
however this would force $x_k = x_{j-i+k}$ for all $k = 0, 1, \dots, p-1$, giving rise to a new period $j-i < p$, a contradiction! So any string of length $p$ must have $m(U) \le 1$, so only one of $aU$ and $bU$ occurs in $W$ (going around the circle once more and adding one more letter, that letter being the last letter of $U$), hence $U$ cannot be FU.
PS. Here $aU$ and $bU$ have length $p+1$, so $m(aU)$ and $m(bU)$ are meaningless.
Another simple but important observation is
(*): since $p > 2^k$, at least one of $m(a)$ and $m(b)$ is greater than $2^{n-1}$.
For each $k = 0, 1, \dots, n-1$, define
$$\Omega_k = \{V | V \text{ is a substring of } W \text{ of length } \le p-1, \text{ and } m(V) > 2^k \}.$$
Then take $U_k \in \Omega_k$ such that its length is the longest among those in $\Omega_k$. By (*), $\Omega_k$ is nonempty, so $U_k$ must exist.
Now fix a $k \in \{0, 1, \dots, n-1\}$. Since $aU_k$ is longer than $U_k$, by the definition we must have $m(aU_k) \le 2^k$; however we also need $2^k < m(U_k) = m(aU_k) + m(bU_k)$ to be satisfied, so $m(bU_k) \ne 0$. By the same reasoning we also get $m(bU_k) \ne 0$. Applying the same reasoning again we get $m(U_ka), m(U_kb) \ne 0$. We conclude that $U_k$ is FU.
Next we must show that these $n$ values of $U_k$ are all distinct from each other: if $U_k \in \Omega_{k+1}$, that is $m(U_k) > 2^{k+1}$, then at least one of $m(aU_k)$ and $m(bU_k)$ would be greater than $2^k$. Since $aU_k$ and $bU_k$ are longer than $U_k$, this again contradicts the definition of $U_k$! So we obtain $2^0 < m(U_0) \le 2^1 < m(U_1) \le 2^2 < \dots < 2^{n-1} < m(U_{n-1})$, which means these $n$ values of $U_k$ are all distinct. This completes the proof!