I have only glanced at the proofs for the first part of the Fundamental Theorem of Calculus, as an exercise I tried to prove it myself. I kindly ask you for some feedback on the solidity, validity, etc. of my attempt.
I use the limit of a sequence rather than the continuous limit since in my analysis class we are mostly using the definition $lim_{x\to a} f(x)=lim_{n\to\infty} f(x_{n}), (x_{n})\to a$, where $lim_{n\to\infty} f(x_{n})$ is the same for all sequences $(x_{n})\to a$
Let f:[a,b] $\rightarrow \mathbb{R}$ be continuous (and thus integratable) and $$F(x):= \int_a^x f(t)dt \quad \space p \in [a,b]$$
$$F'(p)=\lim_{x\to p} \dfrac{F(x)-F(p)}{x-p}=\lim_{x\to p} \dfrac{\int_a^{x} f(t)dt-\int_a^p f(t)dt}{x-p}$$
so we have to look at the limit for a general sequence general $(x_{n})_{n \in\mathbb{N}} \to p$, $x_{n}\in[a,b] \space \forall n \in\mathbb{N}$
$$\lim_{n\to \infty} \dfrac{F(x_{n})-F(p)}{x_{n}-p}=\lim_{n \to\infty} \dfrac{\int_a^{x_{n}} f(t)dt-\int_a^p f(t)dt}{x_{n}-p}=\lim_{n\to\infty}\dfrac{\delta(x_{n},p)\int_{min(x_{n},p)}^{max(x_{n},p)} f(t)dt}{x_{n}-p}$$ with $\delta:= \left\{ \begin{array}{lr} 1 & x_{n}\geq p \\ -1 & x_{n}< p \end{array} \right/$
If $x_{n}\geq p$: $$\int_a^{x_{n}} f(t)dt-\int_a^p f(t)dt=\int_{p}^{x_{n}}f(t)dt=\int_{min(x_{n},p)}^{max(x_{n},p)} f(t)dt=\delta\int_{min(x_{n},p)}^{max(x_{n},p)}f(t)dt$$
If $x_n<p$: $$\int_a^{x_{n}} f(t)dt-\int_a^p f(t)dt=-(\int_a^p f(t)dt-\int_a^{x_{n}} f(t)dt)=\delta(\int_{x_{n}}^{p}f(t)dt)=\delta\int_{min(x_{n},p)}^{max(x_{n},p)}f(t)dt$$
$min(x_{n},p),max(x_{n},p) \in [a,b]$ so $f$ is continuous in this interval, so according to the mean value theorem $\exists\theta_{n}\in[min(x_{n},p),max(x_{n},p)]$ with $$f(\theta_{n})(max(x_{n},p)-min(x_{n},p))=\int_{min(x_{n},p)}^{max(x_{n},p)}f(t)dt$$
This forms a sequence $(\theta_n)_{n\in\mathbb{N}}$ where $\theta_{n}$ is any $\theta_{n}\in[min(x_{n},p),max(x_{n},p)]$ with $$f(\theta_{n})(max(x_{n},p)-min(x_{n},p))=\int_{min(x_{n},p)}^{max(x_{n},p)}f(t)dt$$
$\lim_{n\to\infty} \dfrac{\delta\int_{min(x_{n},p)}^{max(x_{n},p)} f(t)dt}{x_{n}-p}=\lim_{n\to\infty} \dfrac{\delta f(\theta_{n})(max(x_{n},p)-min(x_{n},p))}{x_{n}-p}=\lim_{n\to\infty} f(\theta_{n})$ since $\delta(max(x_{n},p)-min(x_{n},p))=x_{n}-p$ in all cases
$f$ is continuous in [a,b] and all $\theta_{n} \in [min(x_{n},p),max(x_{n},p)] \subseteq [a,b]$ so $\lim_{n\to\infty} f(\theta_{n})= f(\lim_{n\to\infty} \theta_{n})$
$\dfrac{x_{n}+p-|x_{n}-p|}{2}=min(x_{n},p)\leq\theta_{n}\leq max(x_{n},p)=\dfrac{x_{n}+p+|x_{n}-p|}{2}$
$x_{n}$ is a sequence that converges to p so:
for all $\epsilon >0 \space\exists N:|x_{n}-p|<\epsilon \space \forall n\geq N$
$\\ -\epsilon<x_{n}-p<\epsilon \Rightarrow -\epsilon+2p<x_{n}+p<\epsilon+2p$
$\\\\$ $-\epsilon+p=\dfrac{-\epsilon +2p -\epsilon}{2}\leq\dfrac{x_{n}+p-|x_{n}-p|}{2}\leq\theta_{n}\leq \dfrac{x_{n}+p+|x_{n}-p|}{2}\leq\dfrac{\epsilon +2p +\epsilon}{2}=\epsilon+p$
$-\epsilon<\theta_{n}-p<\epsilon \Rightarrow |\theta_{n}-p|<\epsilon \space \forall n\geq N$
So $F'(p)=f(\lim_{n\to\infty}\theta_{n})=f(p)$
Arguably not an answer, but awkward for a comment yet hopefully helpful as feedback. Two algebraic observations will greatly simplify your approach:
Re-write $$ \frac{\int_{a}^{x} f(t)\, dt - \int_{a}^{p} f(t)\, dt}{x - p} $$ as $$ \frac{1}{x - p} \int_{p}^{x} f(t)\, dt. $$
Write $f(p)$ (N.B. not $F'(p)$) as $$ \frac{1}{x - p} \int_{p}^{x} f(p)\, dt. $$ (Right? :)
Now consider the absolute value of the difference, and use monotonicity of the integral, the triangle inequality, and continuity of $f$ at $p$ to show $$ |F'(p) - f(p)| = \lim_{x \to p} \biggl|\frac{1}{x - p} \int_{p}^{x} \bigl(f(t) - f(p)\bigr)\, dt\biggr| = 0. $$