Here's a useful topological fact.
Fact. If $K$ is a compact subset of a domain $\Omega$, then there exists a domain $U$ such that $K\subset U$ and $\overline{U}$ is a compact subset of $\Omega$.
Proof: Fix a point $z_0\in \Omega$. For $n=1,2,\dots$ let $G_n$ be the connected component of $z_0$ in the set $$\{z\in\Omega: |z|<n, \ \operatorname{dist}(z,\partial \Omega)>1/n\}$$
Using the connectedness of $\Omega$, one can show that $\bigcup_n G_n=\Omega$. Therefore, the sets $G_n$ form an open cover of $K$. There is a finite subcover; since the sets $G_n$ are nested, this means $K\subset G_n$ for some $n$. This $G_n$ is the desired $U$. $\quad\Box$
Back to the issue. It's not really about holomorphic functions and their derivatives. We are to prove the following:
Theorem 1. If statement $A$ holds on every compact subset of domain $\Omega$, then statement $B$ holds on every compact subset of $\Omega$.
But instead we prove
Theorem 2. If statement $A$ holds on domain $\Omega$, then statement $B$ holds on every compact subset of $\Omega$.
Indeed, suppose Theorem 2 is proved. Given $\Omega$ as in Theorem 1 and its compact set $K$, take $U$ from the topological fact. Since $\overline{U}$ is a compact subset of $\Omega$, property $A$ holds on $U$. Theorem 2 says that property $B$ holds on $K$, which was to be proved.
Try this: without the accumulation point assumption, it is possible that subsequences of $f_n$ can converge to different limit functions. However, if these limit functions are holomorphic and agree on a set with an accumulation point, then what do you know about them?
Best Answer
By the maximum modulus principle, we have
$$ \sup_{z \in \overline{\Omega}} |f_n(z) - f_m(z)| = \sup_{z \in \partial \Omega} |f_n(z) - f_m(z)|. $$
Since $f_n$ converges uniformly on $\partial \Omega$ to some function, the sequence $(f_n)$ is uniformly Cauchy on $\partial \Omega$ so the right hand side tends to zero as $m,n \to \infty$. This implies that $(f_n)$ is uniformly Cauchy on $\overline{\Omega}$ so $f_n$ converge uniformly to some continuous function $f$ on $\overline{\Omega}$. In particular, $f_n$ also converges uniformly to $f$ on every compact subset of $\Omega$ so $f$ is holomorphic on $\Omega$.