While reading Conway's GTM11, I encountered the following theorem. In the theorem, $G\subset\mathbb{C}$ is a domain and $H(G)$ means the holomorphic functions on $G$, and $C(G,\mathbb{C})$ means the continuous functions on $G$.
(Chap VII.) 2.1 Theorem. If $\{f_n\}$ is a sequence in $H(G)$ and $f$ belongs to $C(G,\mathbb{C})$ such that $f_n\to f$ then $f$ is analytic and $f_n^{(k)}\to f^{(k)}$ for each integer $k\geq1$.
And the proof says
Proof. We will show that $f$ is analytic by Morera's theorem. Let $T$ be a triangle contained inside a disk $D\subset G$. Since $T$ is compact, $\{f_n\}$ converges to $f$ uniformly over $T$. Hence...
My question: how does it follow from $T$ being compact that $f_n\to f$ is uniform?
From $T$ being compact I can conclude that there exists $0\leq\delta_n<\infty$ such that $|f_n(z)-f(z)|\leq\delta_n$ for all $z\in T$. If I can show that $\delta_n\to0$, then the convergence is uniform. So I tried to do this by contradiction. Suppose $\delta_n\not\to0$, and there exists $\delta>0$ and $\{z_{n_k}\}\subset T$ such that
$$|f_{n_k}(z_{n_k})-f(z_{n_k})|>\delta>0$$
Since $T$ is compact and $z_{n_k}\in T$, $\{z_{n_k}\}$ has a convergent subsequence, hence we may assume $\{z_{n_k}\}$ is convergent in the first place and denote the putative limit by $z_0$. Then we can write
$$0<\delta<|f_{n_k}(z_{n_k})-f(z_{n_k})|\\ \leq|f_{n_k}(z_{n_k})-f_{n_k}(z_0)|+|f_{n_k}(z_0)-f(z_0)|+|f(z_0)-f(z_{n_k})|$$
Obviously the second and the third term can be made arbitrarily small. And if I could show the first term can also be arbitrarily small, I would obtain a contradiction and finish the proof. But I don't how whether or how this can be done. (I do know that if $\{f_n\}$ is equicontinuous over $T$ then we are done, but it is not in the assumption of the theorem)
Any suggestions? Thanks in advance.
The above Theorem 2.1 is a famous theorem of Weierstraß. In this theorem $\to$ means uniform convergence on compact subsets.