No: if $\Omega:=\{z,|z|<1\}$, and $f_n(z):=z^n$, this sequence converges uniformly to $0$ on compact sets (because such a set is contained in $B(0,r),r<1$) but not on $\Omega$ as $f_n(1-n^{—1)}\to e$.
You cannot prove that, since it does not necessarily hold.
A simple counterexample is $U = \mathbb{C}$, and $f_n(z) = z^n$, with $z_0 = 0$.
We have $\lim\limits_{n\to\infty} f^{(k)}_n(z_0) = 0$ for all $k \geqslant 0$, but $(f_n)$ converges locally uniformly only in the open unit disk, not in all of $U$.
Choose $f_n(z) = n^n\cdot z^n$, and you don't even have locally uniform convergence in any neighbourhood of $z_0$.
If there is an additional hypothesis that $\{f_n : n \in \mathbb{N}\}$ is a normal family, say the sequence is locally bounded, things are of course different.
Then by normality, $(f_n)$ has a locally uniformly convergent subsequence $(f_{n_{k}})$. Let $f$ be the limit of this subsequence. It remains to see that the full sequence converges locally uniformly to $f$. Again by normality, every subsequence $(f_{n_m})$ has a locally uniformly convergent subsequence $\bigl(f_{n_{m_r}}\bigr)$. Let $g$ denote its limit. Since $f_n^{(k)}(z_0) \to f^{(k)}(z_0)$ for all $k$, and by Weierstraß' theorem $f_{n_{m_r}}^{(k)}(z_0) \to g^{(k)}(z_0)$, we have $f^{(k)}(z_0) = g^{(k)}(z_0)$ for all $k$, hence by the identity theorem, $g \equiv f$.
Thus every subsequence of $(f_n)$ has a subsequence converging locally uniformly to $f$, and that implies that the full sequence converges locally uniformly to $f$.
Best Answer
By Montel's theorem, the sequence $(f_n)$ has a subsequence which converges locally uniformly to some analytic function $f: U \to \mathbb{C}$. Assume, $f_n \not \to f$ locally uniformly. Then there exists a compact set $K$, $\varepsilon>0$ and a subsequence $(f_{n_k})_k$ of $(f_n)$ such that $$\forall k: \|f_{n_k}-f\|_K \geq \varepsilon \tag{1}$$
By Montel's theorem, $(f_{n_k})$ has a locally uniformly convergent subsequence $(f_{n_{k_j}})_j$, $$f_{n_{k_j}} \to \tilde{f} \quad \text{locally uniformly}$$ By assumption, $\tilde{f}|_D = f|_D$. Since $f$, $\tilde{f}$ are holomorphic and $D$ has an accumulation point, we conclude $f = \tilde{f}$ (by the identity theorem). Contradiction to (1)!