I don't know if this will answer all your questions, but what the author is doing is to assume there is a non-zero polynomial which evaluates to zero at all points, then reach a contradiction. He does this by analyzing the hypothetical polynomial in a number of different ways:
Assume the polynomial $p(x)$ evaluates to $0$ at all points, but is not equal to zero. Then it has some definite, positive degree $n$. Since $p(0) = 0$, we must have $p(x) = x\cdot q_0(x)$ for some non-zero polynomial $q_0$.
Since $p(1) = 0$, we must have $q_0(1) = 0$, so $q_0(x) = (x-1)q_1(x)$. Therefore we must have $p(x) = x(x-1)q_1(x)$ for some non-zero polynomial $q_1$.
Now he continues this until he reaches $p(x) = x(x-1)(x-2)(x-3)\cdots (x-n + 1)(x-n)q_n(x)$ for some polynomial $q_n$. But the left side has degree $n$, while the right side has degree at least $(n+1)$, which is a contradiction.
The linear equation he mentions goes like this:
Assume $p(x)$ has positive degree $n$. I'm going to illustrate with $n = 2$, but it's the same for all positive integers. We can then write $p(x) = ax^2 + bx + c$. To find $a, b$ and $c$, we make use of the following: $ 0 = p(0) = a\cdot 0^2 + b\cdot 0 + c = c \quad\Longrightarrow \quad c=0 $ as well as $ 0 = p(1) = a\cdot 1^2 + b\cdot 1 + c = a + b\quad\Longrightarrow \quad a = -b $ and $ 0 = p(2) = a\cdot 2^2 + b\cdot 2 + c = 4a + 2b \quad\Longrightarrow \quad 2a = -b $
to which the only solutions are $a = b = c = 0$, and $p$ is the zero polynomial.
The last one he mentions, Taylor's theorem, depends on the fact that many functions are completely determined by its value and all its derivatives at a single point. Polynomials are among these functions. Since $p$ always evaluates to zero, all its derivatives do too, and the only Taylor series which satisfies this is the Taylor series for the zero function.
Edit: Somehow I completely misread the first paragraph. What he does here is a variation on the first explanation I gave, only he proves that all the coefficients has to be zero the following way:
Assume $p(x) = a_nx^n + \cdots + a_1x + a_0$. Since $p(0) = 0$, we know that $a_0 = 0$. Now examine the function $a_nx^{n-1} + \ldots + a_2x + a_1 = \frac{p(x)}{x}$, and suppose $a_1 \neq 0$. Then the inequalities listed in the text arrives at a contradiction, so $a_1 = 0$. Now proceed similarily with $\frac{p(x)}{x^2} = a_nx^{n-2} + \ldots + a_3x + a_2$, and $a_2$ has to be zero. And so on.