Your idea is fine, and it is possible to make it rigorous using the strong Markov property of Brownian motion. However, it's a bit involved (at least as far as I see) because we need to define iteratively suitable stopping times. Here is a slightly different approach which seems simpler to me:
Define
$$\tau_k := \inf\{t>0; B_{t+\tau_0+\ldots+\tau_{k-1}}=0\}, \qquad \tau_0=0,$$
and
$$B_t^{(k)} := B_{t+\tau_0+\ldots+\tau_{k-1}}, \qquad t \geq 0, \, k \geq 1.$$
By the strong Markov property of Brownian motion, the process $(B_t^{(k)})_{t \geq 0}$ is a Brownian motion for every $k \geq 1$. If we set $\sigma_k := \sum_{i=0}^k \tau_i$, then
\begin{align*} \int_0^{\infty} \sin^2(B_s) \, ds &= \sum_{k=0}^{\infty} \int_{\sigma_k}^{\sigma_{k+1}} \sin^2(B_s) \, ds \\ &= \sum_{k=0}^{\infty} \underbrace{\int_0^{\tau_{k+1}} \sin^2(B_s^{(k)}) \, ds}_{=:X_k}. \end{align*}
Since each $B^{(k)}$ is a Brownian motion, the random variables $X_k$ are identically distributed. Moreover, the strong Markov property yields that $X_k$, $k \geq 1$, are independent. The stopping time $\tau_1$ is strictly positive with probability $1$, and this implies that there exists a constant $c>0$ such that
$$\mathbb{P}(X_1 \geq c)> \frac{1}{2},$$
and hence
$$\mathbb{P}(X_k \geq c)> \frac{1}{2}, \qquad k \geq 1.$$
Consequently,
$$\sum_{k \geq 1} \mathbb{P}(X_k \geq c) =\infty,$$
which implies by the Borel Cantelli lemma that $X_k \geq c$ infinitely often with probability $1$. Thus, $\sum_{k=1}^{\infty} X_k=\infty$ almost surely.
Remark: The above reasoning shows, more generally, that
$$\int_{(0,\infty)} f(B_s) \, ds = \infty \quad \text{a.s.}$$
for every non-negative, measurable function $f$ which is strictly positive on some interval $(0,\epsilon)$. Integrals of the form $\int_{(0,\infty)} f(B_s) \, ds$ are called perpetual integrals.
