1. Assume the contrary: Suppose that after some index $k$, all terms $a_{k+i}$ are composite numbers. This means there exists an index $k$ such that for all $i \geq 1$, $a_{k+i}$ is composite.

2. **Sum starting from $a_k$**: Since the sum
$$\frac{1}{\sqrt{a_1a_2}} + \frac{1}{\sqrt{a_2a_3}} + \frac{1}{\sqrt{a_3a_4}} + \cdots$$
is unbounded, the sum starting from $a_k$ should also be unbounded. Let's denote this sum by $T$:
$$T = \frac{1}{\sqrt{a_k a_{k+1}}} + \frac{1}{\sqrt{a_{k+1} a_{k+2}}} + \frac{1}{\sqrt{a_{k+2} a_{k+3}}} + \cdots$$

3. **Bounding the sum $T$**: Since all $a_{k+i}$ are composite, each $a_{k+i}$ has at least two prime divisors. Therefore, we can write:
$$a_{k+i} \geq p_1 p_2$$
where $p_1$ and $p_2$ are the smallest prime factors of $a_{k+i}$. This implies:
$$\sqrt{a_{k+i} a_{k+i+1}} \geq \sqrt{(p_1 p_2)(p_3 p_4)} \geq \sqrt{2 \cdot 3 \cdot 2 \cdot 3} = 6$$
Hence,
$$\frac{1}{\sqrt{a_{k+i} a_{k+i+1}}} \leq \frac{1}{6}$$

4. Comparing with a known series: Consider the series:
$$\frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \cdots$$
This series can be written as:
$$\sum_{n=2}^{\infty} \frac{1}{n(n+1)}$$
Using partial fractions, we have:
$$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$
Therefore, the series telescopes:
$$\sum_{n=2}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \cdots = \frac{1}{2}$$

5. **Bounding $T$**: Since $\frac{1}{\sqrt{a_{k+i} a_{k+i+1}}} \leq \frac{1}{6}$, we have:
$$T < \sum_{n=2}^{\infty} \frac{1}{n(n+1)} = \frac{1}{2}$$
This implies that $T$ is bounded by $\frac{1}{2}$.

6. Contradiction: The assumption that $T$ is bounded contradicts the given condition that the sum is unbounded. Therefore, our initial assumption that all $a_{k+i}$ are composite must be false.

7. Conclusion: Hence, the sequence $(a_n)$ must contain infinitely many primes.

$\blacksquare$