Showing that the derivative of the integral is the function itself The task at hand is to show that the integral for some function $f$ obeying $\oint f(z) dz = 0$,
\begin{align*}
F(z) = \int_a^{z} f(w) dw,
\end{align*}
is well defined, and that it obeys the relation
\begin{align*}
\lim_{h\rightarrow0} \frac{F(z+h) - F(z)}h = f(z), \tag{$\dagger$}
\end{align*}
that is, the derivative of the integral is well defined to be the function itself.
Assuming I've already proven that the integral is well-defined (i.e. it is unique and it exists), I have to show that the derivative of the integral is the function itself (expression $(\dagger)$). I've managed to formulate a proof which I just handed in as "incomplete, but whatever" since I couldn't figure out what step to take next. 
Now here comes the kicker, my TA graded the proof as sufficient and I'm puzzled as to why. Therefore my question is: why is my proof sufficient? Should I argue that I can keep choosing $w-z$ to be smaller so that the limit goes to zero? What's needed to understand my own proof (I know how silly this sounds...).

First I calculate what the function $f(z)$ integrated over variable $w$ is from $z$ to $z+h$ (why we're doing this will be apparent later)
\begin{align*}
\int_z^{z+h} f(z) dw = f(z) \int_z^{z+h} dw = h f(z) \Rightarrow f(z) = \frac{\int_z^{z+h} f(z) dw}{h}
\end{align*}
and calculate
\begin{align*}
\lim_{h\rightarrow 0} \frac{F(z+h) - F(z)}h &= \lim_{h\rightarrow 0} \frac{\int_a^{z+h} f(w)dw - \int_a^{z} f(w)dw}h\\
&= \lim_{h\rightarrow 0} \frac{\int^{z+h}_{z} f(w)dw+\int_a^{z} f(w)dw - \int_a^{z} f(w)dw}h\\
&= \lim_{h\rightarrow 0} \frac{\int_z^{z+h} f(w)dw}h.
\end{align*}
Now conclude
\begin{align*}
\lim_{h\rightarrow 0} \frac{F(z+h) - F(z)}h -f(z) &= \lim_{h\rightarrow 0} \frac{\int_z^{z+h} [f(w)-f(z)] dw}h.
\end{align*}
If we now use the ML inequality, it follows that
\begin{align*}
\left\lvert \frac{\int_z^{z+h} [f(w)-f(z)] dw}h\right\rvert = \frac1{\lvert h\rvert} \left\lvert\int_z^{z+h} [f(w)-f(z)] \right\rvert dw \leq \frac1{\lvert h\rvert} \varepsilon\lvert h\rvert = \varepsilon,
\end{align*}
for some constant $\varepsilon$, obeying $\lvert f(w)-f(z)\rvert <\varepsilon$ when $\lvert w-z\rvert <\delta$, $\delta>0$.
 A: There's an important piece of information missing here, you don't say what is assumed of the domain of $f$ [you also don't mention that $f$ is assumed to be continuous, but that's inferable from the context alone].
If the domain of $f$ is assumed to be convex (and nonempty, to avoid trivialities), then the only point is that at the end you write "$\lvert \dotsc \rvert \leq \varepsilon$ for some constant $\varepsilon$, obeying $\lvert f(w) - f(z)\rvert < \varepsilon$ when $\lvert w-z\rvert < \delta$, $\delta > 0$". The correct order is that you first pick an arbitrary $\varepsilon > 0$, and then deduce that there is a $\delta > 0$ such that
$$0 < \lvert h\rvert < \delta \implies \biggl\lvert \frac{F(z+h) - F(z)}{h} - f(z)\biggr\rvert \leqslant \varepsilon.\tag{1}$$
By the continuity of $f$, for the given $\varepsilon$ there is an $\eta > 0$ such that $\lvert w-z\rvert < \eta$ implies $\lvert f(w) - f(z)\rvert < \varepsilon$, and choosing $0 < \delta \leqslant \eta$ so that $\lvert h\rvert < \delta$ guarantees that the straight line segment connecting $z$ and $z+h$ lies in the domain of $f$ then yields $(1)$ following your computation - of course for a convex domain, the latter condition is trivially satisfied, so one chooses $\delta = \eta$ (and doesn't even mention $\eta$). One can also put that informally at the end, writing something like "$\lvert\dotsc\rvert \leqslant \varepsilon$ for any given $\varepsilon > 0$ if $0 < \lvert h\rvert < \delta$ for an appropriate $\delta > 0$". In a complex analysis course, one might assume that it is known what "appropriate" means here, and that it can be correctly translated into [more] formal language. But it may be necessary to make the "appropriate" more explicit if one can't count on that assumption.
If the domain of $f$ is not assumed to be convex, there are some more small explanations missing, and your TA has been a bit magnanimous if they awarded full marks for it (the essential bits are there, so I wouldn't deduct much, but without a convexity assumption I'd have deducted some points).
First thing is that it's not said what $\int_a^z f(w)\,dw$ means. If the domain of $f$ is convex, or just star-shaped with respect to $a$, we have the standard interpretation that it should be the integral over the straight line segment connecting $a$ and $z$, and no explanatory remark is necessary. If the domain of $f$ is an arbitrary (nonempty) connected open set, it ought to be said that one chooses a path $\gamma$ (sufficiently regular for integration, that could mean piecewise $C^1$; rectifiable would suffice) from $a$ to $z$ and lets $\int_a^z f(w)\,dw$ denote $\int_{\gamma} f(w)\,dw$. Since the value of the integral is independent from the chosen path (as was shown before), the short-hand of dropping the path and mentioning only start and end points is legitimate.
Next, in your calculation you split $\int_a^{z+h} f(w)\,dw$ into $\int_z^{z+h} f(w)\,dw + \int_a^z f(w)\,dw$. The splitting is legitimate by the premise that the integral of $f$ over all closed paths shall vanish, and if the domain of $f$ is convex, there is no problem, as we can assume the standard interpretation of integrating over a straight line segment for the integrals. If the domain is not convex, the question is about the interpretation of $\int_z^{z+h} f(w)\,dw$. If that is to mean the integration over an arbitrary path from $z$ to $z+h$, that is valid, but doesn't help in the end. To reach the conclusion (by way of the ML inequality) we need $\int_z^{z+h} f(w)\,dw$ to mean the integral over the straight line segment connecting $z$ and $z+h$ (strictly, we can also take some other family of paths, but we need that the length of the path is bounded by $C\cdot \lvert h\rvert$ for some constant $C$, taking the straight line segment is most convenient and easy). So, without a convexity assumption, you should - at about that place - say that from now on you assume $\lvert h\rvert$ small enough that the straight line segment from $z$ to $z+h$ lies in the domain of $f$. (The domain is open, so there's an $r > 0$ such that the disk $D_r(z) = \{ w \in \mathbb{C} : \lvert w-z\rvert < r\}$ lies in the domain of $f$, then $\lvert h\rvert < r$ guarantees that.)
To recapitulate: First one picks an $r > 0$ so that for $\lvert h\rvert < r$, the straight line segment from $z$ to $z+h$ is contained in the domain of $f$. Then one uses the premise "$\oint f(z)\,dz = 0$" to see that - for $\lvert h\rvert < r$ - one has
$$F(z+h) - F(z) = \int_z^{z+h} f(w)\,dw,$$
where the integral means the integral over the straight line segment. Finally, one uses the continuity of $f$ at $z$ to have for every $\varepsilon > 0$ a $0 < \delta \leqslant r$ with $\lvert w-z\rvert < \delta \implies \lvert f(w) - f(z)\rvert < \varepsilon$. The ML inequality then shows that for $0 < \lvert h\rvert < \delta$ we have
$$\biggl\lvert \frac{F(z+h)-F(z)}{h} - f(z)\biggr\rvert < \varepsilon.$$
