Suppose there are only finitely many composite numbers. Let $c$ be larger than all composite numbers in the sequence. Well-known fact: there exist $c+1$ consecutive, sufficiently large integers that are all composite. Suppose they are $N$, $N+1$, ..., $N+c$. None of these $c+1$ numbers can appear in the sequence. Suppose the largest term smaller than $N$ is $a_k$. $a_{k+1}$ is the sum of two terms. Case 1: It is the sum of two primes. The sum of two prime numbers is either $\le a_k + 2$ or divisible by 2. Then, it is composite. Case 2: It is the sum of a prime and a composite. Then, $N\le a_{k+1}\le a_k + c \le N+c$. It is composite. Case 3: It is the sum of two composites. Then, it is too small. Every case leads to contradiction. QED

To begin with, let's start with a logical and convenient reformulation of the problem. Suppose the opposite — then for some \( M_0 \) and all \( n > M_0 \), the number \( a_n \) is prime. Considering that \( a_3 > 2 \), for \( M = \max(M_0, k, 2) \) and all \( n > M \), it follows that \( a_n \) is prime and that \( a_n = a_{n_1} + a_{n_2} \) for some \( 1 \leq n_1, n_2 \leq n-1 \). The next logical step is as follows — if \( n \) is very large, then at least one of \( n_1 \) or \( n_2 \) must be greater than \( M \). Formally, from now on, let us work only with \( n > 2a_{M} \) — clearly, \( 2a_M \leq a_{2a_M} < a_n = a_{n_1} + a_{n_2} \), and with \( n_1, n_2 \leq M \), we get a contradiction. If we assume that \( n_1, n_2 > M \), then \( a_{n_1} \) and \( a_{n_2} \) are prime (and in particular, odd) numbers, and \( a_n = a_{n_1} + a_{n_2} \) would be even (and thus composite), a contradiction. Therefore: For \( n > 2a_{M}: \) \( a_n \) is prime, and \( a_n = a_{n_1} + a_{n_2} \), where \( 1 \leq n_2 \leq M < n_1 \), and \( a_{n_1} \) is a prime number. In particular, any two terms with indices greater than \( 2a_M \) must both be prime and have a difference not exceeding \( a_M \). From this, it follows more generally that any set of \( a_M + 1 \) consecutive natural numbers contains a prime number. The latter is false due to \( (a_M + 2)! + k \) for \( k = 2, 3, \ldots, a_M + 2 \) (it is divisible by \( k > 1 \) and exceeds \( k \)), a contradiction.