If $a_2=0$, there is nothing to prove, hence from now on we assume that $a_2\ne0$.
Assume first that $a_n=0$ for every $n\geqslant3$ (as @copper.hat suggested). Then the hypothesis is that $|a_1+a_2x|\geqslant|a_1|-2|x|$ for every $|x|$ small enough, say $|x|\leqslant x_0$. Choose $x$ such that $x\ne0$, $|x|\leqslant x_0$, $a_1a_2x\leqslant0$, and $|a_2|\,|x|\leqslant |a_1|$. Then $|a_1|-|a_2|\,|x|=|a_1+a_2x|\geqslant|a_1|-2|x|$ and $x\ne0$, hence $|a_2|\leqslant2$.
In the general case, consider $A(x)=a_2+\sum\limits_{n=3}^{+\infty}a_nx^{n-2}$. Since the series $\sum\limits_na_nx^n$ has a positive radius of convergence, the function $A$ is continuous at $0$ and $A(0)=a_2\ne0$, hence $A(x)\ne0$, $|A(x)|\leqslant2|a_2|$, and $A(x)$ has the sign of $a_2$, for every $|x|$ small enough, say $|x|\leqslant x_0$. One can, and we will, assume without loss of generality that, furthermore, $|a_1+A(x)x|\geqslant|a_1|-2|x|$ for every $|x|\leqslant x_0$. Choose $x$ such that $x\ne0$, $|x|\leqslant x_0$, $a_1A(x)x\leqslant0$ and $2|a_2|\,|x|\leqslant |a_1|$. Then $|a_1|-|A(x)|\,|x|=|a_1+A(x)x|\geqslant|a_1|-2|x|$ and $x\ne0$, hence $|A(x)|\leqslant2$. One can choose $x$ with all these properties as close to $0$ as desired, hence, by the continuity of $A$ at $0$, $|a_2|\leqslant2$.