Prove: If $f'(a)=0$, $f'\gt 0$ in some interval left of $a$, and $f'<0$ in some interval right of $a$, then $x$ is a local maximum point. In Spivak's Calculus, the Chapter 11 exposition makes the following claim without a formal proof:

If $a$ is a critical point of $f$ (i.e. $f'(a)=0$), $f'\gt 0$ in some interval to the left of $a$, and $f'<0$ in some interval to the right of $a$, then $a$ is a local maximum point.

In this book, a local maximum point is defined as:

Let $f$ be a function and $A$ a set of numbers contained in the domain of $f$. A point $a$ in $A$ is a local maximum [minimum] point for $f$ on $A$ if there is some $\delta \gt 0$ such that $a$ is a maximum [minimum] point for $f$ on $A \ \cap \ (a-\delta,  a+\delta)$

Spivak then provides the classic picture of an upside down parabolic-shaped segment where the point $a$ is drawn as the maximum point and $f(a)$ the maximum value attained along the parabolic curve (in order to show that the above claim has validity).
Without referencing Spivak's illustration, it is not immediately apparent to me that the point $a$ is, in fact, a local maximum point. In this post, I want to prove Spivak's claim by demonstrating that, given the assumptions:

$f(a)$ is the least upper bound in some neighborhood around $a$.

I think the logic is correct, but I wanted to make sure I did not overlook anything...in particular, how I defined my intervals.

Let $I_L$ and $I_R$ represent the left and right intervals, respectively, where $f'$ at any $x \in I_L$ is greater than $0$ and for any $x \in I_R$, $f'(x)<0$. $I_L$ can be viewed as an open interval of the form $(a-\delta_L, a)$ and $I_R$ viewed as $(a, a+\delta_R)$, for $\delta_L,\delta_R \gt 0$. The argument will work for $I_L= [a-\delta_L, a)$ and $I_R=(a, a+\delta_R]$ as well. It follows that $f$ is strictly increasing on $I_L$ and that $f$ is strictly decreasing on $I_R$.  Next, because $f$ is differentiable at $a$, we know that $f$ is continuous at $a$.
Consider the following set $S=\left\{s: s \in I_L \cup I_R \cup \{a\} \right\}$, or equivalently $S=\left\{s: s \in (a-\delta_L,a+\delta_R)\right\}$. Next, consider the set $T=\{f(s):s\in S\}$. To show that $f(a)$ is the least upper bound of $T$, we need to first show that it is an upper bound of $T$ and then second show that for any $\varepsilon \gt 0$, there is some $t \in T$ such that $f(a)-t \lt \varepsilon$.

Prove: $f(a)$ is an upper bound of $T$ - show that for any $t \in T$, $f(a) \geq t$
Let $x^*$ be an arbitrary element in $S$. If $x^*=a$, then clearly $f(a) \geq f(x^*)$. So consider if $x^* \in I_L$ or $x^* \in I_R$. Assume the former.
By trichotomy of $\mathbb R$, $f(x^*) \lt f(a)$ or $f(x^*) = f(a)$ or $f(x^*) \gt f(a)$. Suppose $f(x^*) \gt f(a)$. Consider  a $\varepsilon^*=f(x^*)-f(a)\gt 0$. By continuity of $f$ at $a$, we know that $f$ is left continuous at $a$. Therefore, we have some $\delta_{\varepsilon^*}$ such that for any $z$: $0 \leq a-z \lt \delta_{\varepsilon^*} \rightarrow |f(z) -f(a)| \lt \varepsilon^*= f(x^*)-f(a)$.
Next, consider the distance $\delta'=\frac{|a-x^*|}{2}$ and the corresponding interval $(a-\delta',a)$. Notice that $(a-\delta',a) \subset I_L$.  Let $\delta^*=\min(\delta_{\varepsilon^*},\delta')$. Then we can say that for any $z$: $0 \leq a-z \lt \delta^* \rightarrow |f(z)-f(a)|\lt f(x^*)-f(a)$. Choosing an arbitrary $z^*$ that satisfies these conditions, we have: $|f(z^*)-f(a)| \lt f(x^*)-f(a) \implies f(z^*)\lt f(x^*)$. However, our definition of $\delta^*$ (because of $\delta'$) forces $z^* \gt x^*$, but $z^* \in I_L$, which, by assumption of a strictly increasing interval, means $f(z^*) \gt f(x^*)$. This is a contradiction. A similar argument that uses right continuity at $a$ where $x^* \in I_R$ will show that if we assume $f(x^*) \gt f(a)$, a contradiction will arise. Therefore, for any $s \in S$, we have that $f(s) \leq f(a)$, which means that $f(a)$ is an upper bound of $T$.

Prove: $f(a)$ is the least upper bound of $T$ - show that for any $\varepsilon \gt 0$, there is some $t \in T$ such that $f(a)-t \lt \varepsilon$.
By continuity of $f$ at $a$, we know that for any $\varepsilon$, there is a corresponding $\delta_{\varepsilon}$ such that for any $x$: $0 \leq |x-a| \lt \delta_{\varepsilon} \rightarrow |f(a)-f(x)|\lt \varepsilon$. Recall our definitions of $I_L$ and $I_R$, which made use of $\delta_L$ and $\delta_R$, respectively. Consider $\delta_S=\min(\delta_L, \delta_R)$. Next let $\delta=\min(\delta_{\varepsilon},\delta_S)$.
From continuity and our first proof, we know that for any $x$: $0 \leq |x-a| \lt \delta \rightarrow |f(a)-f(x)|=f(a)-f(x)\lt \varepsilon$. By our definition of $\delta$, these $x$ are guaranteed to be within $S$ (which means any corresponding $f(x)$ is an element of $T$) and are guaranteed to satisfy the condition that $f(a)-f(x) \lt \varepsilon$. Therefore, choosing any such $x$ within this interval will do the trick, which means that $f(a)$ is the least upper bound of $T$.

Placing all of this in the context of the definition of "local maximum point", let $A = \text{dom}(f)$ and let $\delta=\min(\delta_L,\delta_R)$. Then $a$ is a local maximum point for $f$ on $A \cap (a-\delta, a+\delta)$.
 A: 
If $a$ is a critical point of $f$ (i.e. $f'(a)=0$), $f'>0$ in some interval to the left of $a$, and $f'<0$ in some interval to the right of $a$, then $a$ is a local maximum point.

Note first, I've corrected Spivak's typo (your quote says "then $x$[sic] is a local maximum point."
Here's a more direct proof using the MVT, as sketched out in my comment.
Let $x$ be any point in the left interval.
Applying the MVT we see
$$\frac{f(a) - f(x)}{a-x} = f'(x_0),$$
where $x_0$ is a point between $x$ and $a$.
Since $x_0$ is also in the left interval, $f'(x_0) > 0$, and we have
$$\frac{f(a) - f(x)}{a-x} > 0.$$
The denominator here is greater than zero. Therefore, we must have
$$f(a) - f(x) > 0,$$
or
$$f(a) > f(x).$$
This applies to all $x$ in the left interval.
A similar argument will show $f(a) > f(x)$ for all $x$ in the right interval.
$$\blacksquare$$
A: Another argument, similar to yours:
Suppose we have
$$ x_L < a < x_R,$$
with $f'(a) = 0$, $f'> 0$ on $(x_L,a)$, and $f'< 0$ on $(a,x_R)$, and $f$ is continuous on $[x_L,x_R]$.
Since $f$ is continuous on $[x_L,x_R]$, it is bound on this interval, and there exists some point $y$ in $[x_L,x_R]$ such that $f(y) \geq f(x)$ for all $x$ on $[x_L,x_R]$.
Suppose $y$, this maximum point, is in the left interval. Since $f$ is increasing to the right of $y$, immediately we have a contradiction, with $f(y) < f(x)$ for all $x$ on $(y,a)$.
In other words, $y$ cannot be on the left interval.
Similarly, $y$ cannot be on the right interval, as this is immediately contradicted by points on $(a,y)$.
Therefore, the only remaining possibility is $y = a$.
$\blacksquare$
A: One more alternative view:
We begin just as we did in my second proof:
Suppose we have
$$ x_L < a < x_R,$$
with $f'(a) = 0$, $f'> 0$ on $(x_L,a)$, and $f'< 0$ on $(a,x_R)$, and $f$ is continuous on $[x_L,x_R]$.
Again, since $f$ is continuous on $[x_L,x_R]$, it is bound on this interval, and there exists at least one point $y$ in $[x_L,x_R]$ such that $f(y) \geq f(x)$ for all $x$ on $[x_L,x_R]$.
$y$ is a maximum point of $f$ on $[x_L,x_R]$.
Since $f$ is increasing to the right of $x_L$, this point cannot be a maximum point ($y\neq x_L$.)
Since $f$ is decreasing to the left of $x_R$, this point cannot be a maximum point.
Thus, neither endpoint is a maximum point for $f$ on $[x_L,x_R]$. Therefore, at least one maximum point must be in  $(x_L,x_R)$.
Since $f$ is differentiable everywhere on $(x_L,x_R)$, we know $f'$ exists at all such maximum points, and so $f'= 0$ at all such points. As $a$ is the only such point with $f'= 0$, $a$ must a maximum point for $f$ on $[x_L,x_R]$. Not only that, but $a$ is the only maximum point.
$\blacksquare$
