Sorry if this has been proven previously on MSE but I cannot find an obvious duplicate. I am attempting to prove the stronger version of Morera's theorem namely:
If $f:U\mapsto\mathbb{C}$ is a continuous function on an open set $U$ such that $\int_\gamma f(z)\,\mathrm{d}z=0$ for all triangular contours $\gamma$ contained in $U$, then $f$ is holomorphic on $U$.
Proof (attempt):
Let $a\in U$. Since $U$ is open, $\exists\,r\gt0$ such that $B(a,r)=\{z\in\mathbb{C}:|z-a|\lt r\}\subseteq U$. Now consider $f$ restricted to the domain $B(a,r)$. Using the given assumptions, this restriction of $f$ is continuous and satisfies $\int_\gamma f(z)\,\mathrm{d}z=0$ for all triangular contours $\gamma$ contained in $B(a,r)$. Then we can define $F:B(a,r)\mapsto\mathbb{C}$ by $$F(z)=\int_{[a,z]}f(w)\,\mathrm{d}w$$ where $[a,z]$ is the line segment from $a$ to $z$ in $\mathbb{C}$. This function is now well-defined as $B(a,r)$ is connected. Next we can calculate \begin{align} F'(z) &=\lim_{h\to0}\frac{F(z+h)-F(z)}h\\ &=\lim_{h\to0}\frac{\int_{[a,z+h]}f(w)\,\mathrm{d}w-\int_{[a,z]}f(w)\,\mathrm{d}w}h\\ &=\lim_{h\to0}\frac{\overbrace{\int_{[a,z+h]}f(w)\,\mathrm{d}w+\int_{[z+h,z]}f(w)\,\mathrm{d}w+\int_{[z,a]}f(w)\,\mathrm{d}w}^{=\int_\gamma f(w)\,\mathrm{d}w=0}+\int_{[z,z+h]}f(w)\,\mathrm{d}w}h\\ &=\lim_{h\to0}\frac{\int_{[z,z+h]}f(w)\,\mathrm{d}w}h\\ &=\lim_{h\to0}\frac1h\int_0^1f(z+ht)\cdot h\,\mathrm{d}t\\ &=\lim_{h\to0}\int_0^1f(z+ht)\,\mathrm{d}t\\ &=\int_0^1\lim_{h\to0}f(z+ht)\,\mathrm{d}t\\ &=\int_0^1f(z)\,\mathrm{d}t\qquad(f\text{ continuous})\\ &=f(z)\\ \end{align} Thus $F$ is holomorphic on $B(a,r)$ with derivative $f$. So, in particular, we can apply Cauchy's differentiation formula to give $$f'(a)=F''(a)=\frac1{\pi i}\int_\gamma\frac{F(z)}{(z-a)^3}\mathrm{d}z$$ for a suitable contour $\gamma$. But $a\in U$ was chosen arbitrarily and hence $f$ is holomorphic on $U$.
It looks good. Hope the following adds something.
Regarding the computation of the derivative $F'$, an alternative using triangle inequality: We have from your formulas for $h\neq 0$ $$\dfrac{F(z+h)-F(z)}{h}-f(z)=\frac{1}{h}\int_{[z,z+h]}\{f(\zeta)-f(z)\}d\zeta$$ Now by continuity of $f$, given $\epsilon>0$ there exists $\delta>0$ such that $|\zeta-z|<\delta\Rightarrow |f(\zeta)-f(z)|<\epsilon$ so for $0<|z+h-z|=|h|<\delta$ we get using triangle inequality for contours $$|\dfrac{F(z+h)-F(z)}{h}-f(z)|\le\frac{1}{h} \max{|f(\zeta)-f(z)|}_{\zeta\in [z,z+h]}.h<\epsilon$$
In the end instead of using Cauchy's formula, we can use the fact that derivative of holomorphic is holomorphic (from what I have seen the Cauchy formulas follow from this in composition with the usual formula): We see that $F\in Hol(B(a,r))$ as a result. And so $F'=f\in Hol(B(a,r))$. Now since $a$ was arbitrary we are done.