Spivak's proof that every polynomial of odd degree has a root

The point is not just to ensure that $|1+\sum_{k=1}^{n}a_{n-k}/x^k|$ is positive but that it is actually bounded away from zero, independently of $x$. Actually, the factor $2$ is overkill, but it makes the analysis quick and easy. Harald's comment answers your second question.

Edit: (In response to the OP's query) Yes, it's sufficient to take $x\ge n|a_k|+\varepsilon\;(k=1,\dots,n)$, along with $x>1$, where $\varepsilon>0$ is independent of $x$. Then the multiplying factor $|1+\sum_{k=1}^{n}a_{n-k}/x^k|$ exceeds a constant, which can be written in terms of $\varepsilon$ and the $a_k$, which can be seen to be positive. But why bother when you can just write $\frac12$?