Proof of Runge's theorem

$f+g\in B(E)$:

You have proved that if $f_n\Rightarrow f$ and $g_n\Rightarrow g$ on $K$ then $f_n+g_n\Rightarrow f+g$ on $K$. And it will be better if you state that the pole of $f_n+g_n$ lies in poles of $f_n$ and $g_n$.

There is no need to prove that $f+g$ has a pole in $E$ since $f+g$ is defined on $K$.

$\partial G\subset \partial V$:

Lemma 1. Suppose $X$ is a topological space and $A$ is a component of $X$, then $A$ is closed in $X$.

Lemma 2. Suppose $X$ is an open subset of $\mathbb{C}$ and $A$ is a component of $X$, then $A$ open.

Back to your question, $G$ is open by lemma 2. Let $E=V\backslash G$, then $E$ is open in $V$ by lemma 1 and thus $E$ is open in $\mathbb{C}$.

\begin{align*} \partial V & =\overline{V^c}\cap \overline{V}\\ & =\overline{(G\cup E)^c}\cap \overline{G\cup E}\\ & =(G^c\cap E^c)\cap (\overline{G}\cup\overline{E})\\ & =(G^c\cap E^c\cap \overline{G} )\cup (G^c\cap E^c\cap \overline{E})\\ & =(G^c\cap \overline{G})\cup (E^c\cap \overline{E})\\ & =\partial G\cup \partial E. \end{align*} The penultimate equation is because $\overline{G}\subset E^c$ and $\overline{E}\subset G^c$.

So the following conclusion is obatained:

Conclusion. Suppose $X$ is an open subset of $\mathbb{C}$ and $A$ is a component of $X$, then $\partial X=\partial A\cup\partial (X\backslash A)$.