Without employing Mean Value Theorem, show that $f$ continuous at $a$ and $f'(x)=0$ for all $x\in (a,b)$ means that $f'(a)=0$ Problem 69f in Chapter 11 of Spivak's Calculus reads partially as follows:

Suppose that $f$ is continuous on $[0,1]$ and $f'(a)=0$ for all $a$ in $(0,1)$. Then [prove something]

The solution manual's first line is:

For $\varepsilon \gt 0$, the function $g'(a)=f'(a)+\varepsilon=\varepsilon \gt 0$ for all $a$ in $[0,1]$.

Earlier in this problem, Spivak says to NOT use the Mean Value Theorem (MVT). Given this restriction, I am not sure how one produces the first line of the solution manual. In particular, why does $f'(0)=0$ and why does $f'(1)=0$?
Originally, I would have said $\displaystyle \lim_{x \to 0}f'(x)=0 \rightarrow f'(0)=0$, but this claim (of which the generalization is derived earlier in the book) depends on the MVT, so this must not be how to carry out this proof. Also, earlier in the book we covered Darboux's Theorem...but that, too, required the MVT.
Any thoughts?
We want to show that for any $\varepsilon \gt 0$, there is a $\delta \gt 0$ such that $\forall y \in (0,\delta): \left|\frac{f(y)-f(0)}{y}\right|\lt \varepsilon.$ I feel like some clever $\epsilon-\delta$ arguments can be used in conjunction with $|f(0)-f(x)| \lt \varepsilon_1$ and $\left|\frac{f(y)-f(x)}{y-x}\right|\lt\varepsilon_2$

Spivak, Michael, Calculus. 4th ed, Berkeley, California: Publish or Perish, Inc. (2008). ZBL1272.26002.
 A: I think I have a way of proving that $\displaystyle \lim_{x\to 0^+}\frac{f(x)-f(0)}{x}=0$, but the argument used to prove this statement, in and of itself, would defeat the purpose of even pursuing the statement.
WLG, suppose by contradiction that $\displaystyle \lim_{x\to 0^+}\frac{f(x)-f(0)}{x}=L \gt 0$.
Consider when $\varepsilon_0 = \frac{L}{2}$. Then, for any $x \in (0, \delta_{\varepsilon_0})$, we have $\frac{L}{2} \lt \frac{f(x)-f(0)}{x} \lt \frac{3L}{2}$. Rearranging, we have that $f(0) + \frac{L}{2}x \lt f(x) \lt f(0)+\frac{3L}{2}x \quad (\dagger_1)$.
Referencing the hyperlink (with a slight modification), we will make use of the two functions $g_{\varepsilon}$ and $h_{\varepsilon}$, for any $\varepsilon \gt 0$. For $g_{\varepsilon}$, which has the rule $g_{\varepsilon}(x)=f(x)+\varepsilon (x-a)$, look at the interval $[a,b]$ for any $a \lt b \in (0,1)$. Then $g_{\varepsilon}'(x)=f'(x)+\varepsilon \gt 0$. Therefore, on the interval $[a,b]$, $g_{\varepsilon}$ is strictly increasing. As such, $g(a) \lt g(b)$.
This implies that $f(a)\lt f(b)+ (b-a)\varepsilon$. With some algebra, we have that $\frac{f(a)-f(b)}{a-b} \gt -\varepsilon$. Using $h_{\varepsilon}=(x-a)\varepsilon-f(x)$, we can similarly derive that: $\frac{f(a)-f(b)}{a-b} \lt \varepsilon$.
Together, we then have that, for any $\varepsilon \gt 0: \left|\frac{f(a)-f(b)}{a-b} \right| \lt \varepsilon$, which means $f(a)=f(b)$. $a$ and $b$ were arbitrary, so we can generalize. For any $a,b \in (0,1)$, we have that $f(a)=f(b)$.
Now, consider an arbitrary $x \in (0, \delta_{\varepsilon_0})$. Additionally, consider an $x'=\frac{x}{4}$.
By $(\dagger_1)$ we have that:
$$f(0) + \frac{L}{2}x' \lt f(x') \lt f(0)+\frac{3L}{2}x'$$.
However, note that $\frac{3L}{2}x'=\frac{3L}{8}x \lt \frac{L}{2}x$. Combining these facts, we have:
$$f(x') \lt f(0) + \frac{L}{2}x \lt f(x)$$
As such, there is no way that $f(x')$ can equal $f(x)$, a contradiction.
Therefore, $f'(0)^+ \neq L \gt 0$. A similar argument works when assuming $f'(0)^+ \lt L$. As such, $f'(0)^+=0$.

Given the purpose of this problem, which is to show that $f(0)=f(1)$, the above approach is excessive. Really, once we have shown that for any $(a,b) \in (0,1): f(a)=f(b)$, the assumption of continuity alone at $x=0$ and $x=1$ suffices to show that $f(0)=f(1)$.
