Question on the proof of Theorem 7-1 in Spivak's Calculus

I have a question regarding the following proof in Spivak’s Calculus. (I excerpted the post by Kepler 9 years ago)

Theorem 7-1:

If $f$ is continuous on $[a,b]$ and $f(a) < 0 < f(b)$, then there is some number $z$ in $[a,b]$ such that $f(x) = 0$.

Proof:
Define the set $A$ as follows:

$$A = {x : a le xle b, mbox{ and } f mbox{ is negative on the interval } [a,x] }.$$

     Clearly $A ne emptyset$, since $a$ is in $A$; in fact, there is some $delta > 0$ such that $A$ contains all points $x$ satisfying $a le x < a + delta$; this follows from Problem 6-16, since $f$ is continuous on $[a,b]$ and $f(a)<0$. Similarly, $b$ is an upper bound for $A$ and, in fact, there is a $delta > 0$ such that all points $x$ satisfying $b-delta < x le b$ are upper bounds for $A$; this also follows from Problem 6-16, since $f(b) > 0$.

     
From these remarks it follows that $A$ has a least upper bound $alpha$ and that $a < alpha < b$. We now wish to show that $f(alpha) = 0$, by eliminating the possibilities $f(alpha) < 0$ and $f(alpha) > 0$.

     
Suppose first that $f(alpha) < 0$. By Theorem 6-3, there is a $delta > 0$ such that $f(x) < 0$ for $alpha – delta < x < alpha + delta$. Now there is some number $x_0$ in $A$ which satisfies $alpha – delta < x_0 < alpha$ (because otherwise $alpha$ would not be the least upper bound of $A$). This means that $f$ is negative on the whole interval $[a,x_0]$. But if $x_1$ is a number between $alpha$ and $alpha+delta$, then $f$ is also negative on the whole interval $[x_0,x_1]$. Therefore $f$ is negative on the interval $[a,x_1]$, so $x_1$ is in $A$. But this contradicts the fact that $alpha$ is an upper bound for $A$; our original assumption that $f(alpha) < 0$ must be false.

$$cdots$$

Is "Now there is some number $x_0$ in $A$ which satisfies $alpha – delta < x_0 < alpha$" in the last paragraph above wrong? Should it be $alpha – delta < x_0 le alpha$? Although this won’t affect the proof, I am just wondering whether my understanding of the least upper bound is correct or not. Thanks in advance.