Proof: Let $\mathbb{D}(p,r)=D\subset\Omega$ be a disc. Define a function $F: D\rightarrow\mathbb{C}$ by \[F(z)=\int_{[p,z]} f(\zeta)\,d\zeta\qquad z\in D.\]
Fix some $z_0\in\Omega$. Then if $T=[p,z]\cup [z,z_0]\cup [z_0,p]$ (i.e. the triangle $T$ is made of those three line segments in $D$), we see that
\[\begin{aligned} \int_T f(\zeta)\,d\zeta=0 &\implies \int_{[p,z]} f(\zeta)\,d\zeta + \int_{[z,z_0]} f(\zeta)\,d\zeta + \int_{[z_0,p]} f(\zeta)\,d\zeta = 0\\ &\implies \int_{[p,z]}f(\zeta)\,d\zeta - \int_{[p,z_0]} f(\zeta)\,d\zeta) = \int_{[z_0,z]} f(\zeta)\,d\zeta.\end{aligned}\]
Hence,
\[\frac{F(z)-F(z_0)}{z-z_0}-f(z_0) = \frac{1}{z-z_0} \int_{[z_0,z]}\left( f(\zeta)-f(z_0)\right)\,d\zeta.\]
Since $f$ is continuous at $z_0$, then for each $\varepsilon>0$ there is a $\delta>0$ such that $|f(\zeta)-f(z_0)|<\varepsilon$ whenever $|\zeta-z_0|<\delta$. This now implies that
\[\begin{aligned} \left|\frac{F(z)-F(z_0)}{z-z_0} - f(z_0)\right| &= \left|\frac{1}{z-z_0}\int_{[z_0,z]} \left( f(\zeta)-f(z_0)\right)\,d\zeta\right| \\ &\leq \left|\frac{1}{z-z_0}\right| \int_{[z_0,z]} |f(\zeta)-f(z_0)|\,d\zeta\\ &\leq \varepsilon \left|\frac{1}{z-z_0}\right| \int_{[z_0,z]}\,d\zeta\\ &= \varepsilon\end{aligned}\]
Hence,
\[\left|\frac{F(z)-F(z_0)}{z-z_0} - f(z_0)\right| \leq \varepsilon\qquad \text{if $|z-z_0|<\delta$}\]
It now follows that $F^{\prime}(z_0)$ exists and equals $f(z_0)$. Since $z_0\in D$ was chosen arbitrarily, it follows that $F^{\prime}=f$ (i.e. $f$ has a primitive); furthermore, we have that $F$ is holomorphic in $D$. Since the derivative of a holomorphic function is holomorphic, it now follows that $f$ is holomorphic in $D$. Furthemore, since this is true for every disk $D$ contained in $\Omega$, it must follow that $f$ is holomorphic in $\Omega$.$\hspace{.25in}\blacksquare$