HINT:
You're right: the comparison test is the way to go. In the domain of integration, i.e., the interval $[0,1]$, the only point that causes concern is $0$. As $x \to 0$, note that the integrand grows unbounded. Therefore, to decide the convergence or divergence of the integral, we need to bound the growth of the integrand near $0$. This is the idea behind the (limit) comparison test.
To implement the above idea, we could use the standard fact $\sin x \sim x$ for $x$ close to $0$ (i.e., as $x \to 0$). Therefore, our integral $\int_0^1 \frac{1}{\sqrt{\sin x}} ~\mathrm dx$ converges if and only if $\int_0^1 \frac{1}{\sqrt{x}} ~\mathrm dx$ (the integral of the test function) converges. Do you know how to establish the convergence (or divergence) of the latter integral?
Convergence of the test integral: The integral $\int_0^1 \frac{1}{\sqrt{x}} ~\mathrm dx$ in fact converges (and so does our original integral). To see this, note that $ \int_{\delta}^{1} \frac{1}{\sqrt{x}} ~\mathrm dx = \left. 2 \sqrt{x} \right|_{\delta}^{1} = 2 - 2 \sqrt{\delta} \to 2, $ as $\delta \to 0$.
In fact, one could similarly see that the integral $\int_0^1 x^p ~\mathrm dx$ converges if and only if $p \gt -1$.