Purely by definition, as @Magdiragdag pointed out.
But beware: the subtlety here is that polynomials and polynomials functions are not necessarily the same thing!
Polynomials can be turned into functions if you "plug in" values for X. But if the values for X lies in a finite field then it can happen that different polynomials give rise to the same polynomial function!
For example, take the field $Z_2 = \{0,1\}$ of residues mod 2, then $X^2$ and $X$ are different polynomials in $Z_2[X]$ (they have different coefficients) but the corresponding polynomial functions are the same, since $X^2 = X$ for $X = 0$ and $1$. More generally, the different polynomials $X^p$ and $X$ give rise to the same polynomial function for $X$ lying in the field $Z_p$ of residues mod p (p a prime), as an immediate consequence of Fermat's Little Theorem.
To go on with your argument and conclude that $a_1 = b_1$ in the polynomials seen as polynomial functions over R you must be able to choose a nonzero X in the field R which is not a root of neither $a_n X^{n−1}+\cdots+a_1$, nor $b_n X^{n−1}+\cdots+b_1$, which you can do safely if your ring R is an infinite field, such as the rationals, the reals or the complex numbers.
So, following your proof, for infinite fields R we get that polynomials in R[X] and polynomials functions with domain R are "the same thing". But this is not true for an arbitrary ring R, as the examples above point out.
TL;DR: An essential property of polynomials is that they can be associated to* functions on their coefficient rings. The fact that the evaluation map is a homomorphism is what makes this possible.
In school, one is often taught that polynomials are expressions in which we can "plug in" numbers and "everything works out the same". The evaluation map being a ring homomorphism is how one takes this informal, intuitive understanding of polynomials and makes it rigorous and formal.
*(not necessarily in a bijective manner, see this question)
In some sense, the fact that the evaluation map is a ring homomorphism is the defining property of a polynomial ring. Such a property is also called a universal property.
UNIVERSAL PROPERTY OF POLYNOMIAL RINGS (See also)
In other words, the fact that the evaluation map is a ring homomorphism is the very reason to even bother thinking about polynomial rings in the first place.
The evaluation map being a homomorphism means that it is possible to associate abstract polynomials with actual functions on the ring. If the evaluation map were not a homomorphism, multiplication of polynomials would not correspond to multiplication of their "values" at every point of the ring, and addition of polynomials would not correspond to the addition of their "values" at every point of the ring. Since we expect multiplication and addition of functions to be defined pointwise, the evaluation map being a homomorphism is a necessary condition for the elements of the polynomial ring to correspond to functions on the coefficient ring. (It is also sufficient.)
For example, see this quote from the above-linked page:
The requirement that the rings be commutative is vital. A fundamental difference for polynomials over non-commutative rings is additional difficulty identifying polynomial forms and functions using this method. [emphasis mine]
So perhaps one could say that the book you are reading has the order backwards: instead of defining polynomial rings in terms of the universal property that evaluation maps are ring homomorphisms, they have defined a polynomial ring in terms of a specific construction, and then showed that this construction satisfies the universal property.
This is an understandable approach, because defining something in terms of universal properties may not be a good idea if no structure actually exists which satisfies the universal property.
Still, thinking in terms of universal properties, and subsequently proving that a structure exists which satisfies the universal property, may be more straightforward to some people.
Thinking in terms of universal properties first and constructions second is somewhat akin to declarative programming: we "declare" what properties we want a certain object to have, without worrying as much about the "implementation" (how to construct an object satisfying the property).
The approach used by your textbook is somewhat more like imperative programming: we specify how to construct an object, therefore being assured that it exists, and then retroactively examine what properties this constructed object has.
In either case, we have to worry about both whether an "implementation" exists as well as what properties we want, but in some sense we have the liberty to choose which concern to address first, or at least worry about more, depending upon our preferences and/or personal strengths.
Best Answer
Hint Define $\pi_p: \mathbb{Z}[x] \rightarrow \mathbb{Z}_p[x]$ to be the mod map you describe. It suffices to show that $\pi_p(f+g) = \pi_p(f)+\pi_p(g)$ and $\pi_p(fg) = \pi_p(f)\pi_p(g)$ to prove that $\pi_p$ is a ring homomorphism. This can be done more easily if you know that the mod map taking $\mathbb{Z} \rightarrow \mathbb{Z}_p$ is itself a ring homomorphism.