We can assume $\lim_{n\to\infty} a_n=\infty,$ since otherwise the sequence is a constant from some place on, and this case is trivial. Further, we may assume the sequence is strictly increasing, otherwise we can consider only the distinct terms. If there are $i,j\in\mathbb{N}$ with $(a_i,a_j)=1$ we are done since by Chicken McNugget theorem all sufficiently large $m\in \mathbb{N}$ can be represented as $m=ka_i+\ell a_j, k,\ell\in \mathbb{N}.$ Suppose $(a_i,a_j)>1,\forall i,j\in\mathbb{N}.$ Consider the set $$M:=\{(a_1,a_i) : i\in\mathbb{N}\}$$Since $M$ is a finite set, there exists $N\in\mathbb{N}$ such that $$M=\{(a_1,a_i) : i=1,2,\dots,N\}$$For any $i=1,2,\dots,N,$ again by Chicken McNugget (a variant of it) there exists $m(i)$ such that all $m\ge m(i)$ that are multiple of $(a_1,a_i)$ can be represented as $m=ka_1+\ell a_i.$ Take $m=\max\{m(i):i=1,2,\dots,N\}.$ Now, for any $j$ satisfying $j>N$ and $a_j>m,$ there exists $i, 1\le i\le N$ with $(a_1,a_i)=(a_1,a_j).$ Then $a_j=k a_1+\ell a_i$ for some $k,\ell\in\mathbb{N}.$

Waaat, this is 4th grade???