For any $n \in \mathbb{N}^*$, we have
$$\begin{align*} S_{2^n} &= 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2^n} = 1 + \frac{1}{2} + \left( \frac{1}{2^1} + \frac{1}{2^2} \right) + \\ & \quad \left( \frac{1}{2^{n-1}} + \frac{1}{2^n} \right) \\ &> 1 + \frac{1}{2} + \left( \frac{1}{2^2} + \frac{1}{2^2} \right) + \cdots + \left( \frac{1}{2^n} + \cdots + \frac{1}{2^n} \right) \\ &= 1 + \frac{1}{2} + \frac{1}{2} + \cdots + \frac{1}{2} > \frac{1}{2}n. \end{align*}$$
Let $N_0 = \lfloor \frac{1}{b-a} \rfloor + 1$, $m = \lfloor S_{N_0} \rfloor + 1$. Then $\frac{1}{b-a} < N_0$, $\frac{1}{N_0} < b-a$, and $S_{N_0} < m \le m+a$.
Let $N_1 = 2^{2(m+1)}$. Then $S_{N_1} = S_{2^{2(m+1)}} > m+1 \ge m+b$.
We claim that there exist $n \in \mathbb{N}^*$ with $N_0 < n < N_1$ such that $m+a < S_n < m+b$ (or, in other words, $S_n - [S_n] \in (a, b)$).
Otherwise, assuming the claim is false, then there must exist $k > N_0$ such that $S_{k-1} \le m+a$ and $S_k \ge m+b$.
Then $S_k - S_{k-1} \ge b-a$. But it contradicts the fact that
$$S_k - S_{k-1} = \frac{1}{k} < \frac{1}{N_0} < b - a.$$
Therefore, the claim is true.
Furthermore, assume there are only a finite number of positive integers $n_1, \dots, n_k$ satisfying
$$S_{n_j} - [S_{n_j}] \in (a, b) \quad (1 \le j \le k).$$
Define $c = \min_{1 \le j \le k} \{S_{n_j} - [S_{n_j}]\}$. Then there exists no $n \in \mathbb{N}^*$ such that $S_n - [S_n] \in (a, c)$. It contradicts the above claim.
Therefore, there are infinite terms in the sequence $\{S_n - [S_n]\}$ that are within $(a, b)$.
The proof is complete.

For any $n \in \mathbb{N}^*$, we have
$$\begin{align*} S_{2^n} &= 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2^n} = 1 + \frac{1}{2} + \left( \frac{1}{2^1} + \frac{1}{2^2} \right) + \\ & \quad \left( \frac{1}{2^{n-1}} + 1 + \dots + \frac{1}{2^n} \right) \\ &> 1 + \frac{1}{2} + \left( \frac{1}{2^2} + \frac{1}{2^2} \right) + \dots + \left( \frac{1}{2^n} + \dots + \frac{1}{2^n} \right) \\ &= 1 + \frac{1}{2} + \frac{1}{2} + \dots + \frac{1}{2} > \frac{1}{2}n. \end{align*}$$
Therefore, $S_n$ can be larger than any positive number as long as $n$ becomes sufficiently large.
Let $N_0 = \lfloor \frac{1}{b-a} \rfloor + 1$. Then $\frac{1}{N_0} < b-a$, and when $k > N_0$, we have
$$S_k - S_{k-1} = \frac{1}{k} < \frac{1}{N_0} < b - a.$$
So for any positive integer $m > S_{N_0}$, there exists $n > N_0$ such that $S_n - m \in (a, b)$, or, in other words, $m + a < S_n < m + b$. Otherwise, there must be $k > N_0$ such that $S_{k-1} \le m + a$ and $S_k \ge m + b$, i.e., $S_k - S_{k-1} \ge b - a$. But it contradicts the fact that
$$S_k - S_{k-1} = \frac{1}{k} < \frac{1}{N_0} < b - a.$$
Now let $m_i = [S_{N_0}] + i$ ($i = 1, 2, 3, \dots$). Then there exists $n_i > N_0$ such that $m_i + a < S_{n_i} < m_i + b$, i.e., $S_{n_i} - [S_{n_i}] \in (a, b)$.
Therefore, there are infinite terms in the sequence $\{S_n - [S_n]\}$ that are within $(a, b)$. $\square$