Let $(X,\mathcal{O}$) be a topological space, $(V,\| \cdot \|)$ be a normed vector space, $\mathcal{O}_V$ be the topology induced by $\| \cdot \|$ on $V$.
I want to show, that if $f,g:X \to V$ are two continuous functions, then $f+g:X\to V$ is again continuous.
The definition I'm working with is the following: $f$ is called continuous iff all preimages of open sets are again open.
A long time ago I've seen proofs for the special case $V=\mathbb{K}^n$ and where $\mathcal{O}$ is the topology induced by a metric $d$ on $X$, but couldn't generalize them.
I tried to show this directly by the definition, but got stuck. Here is what I tried:
Let $S=\{B_r(v) \mid r\geq 0, v \in V\}$. Since $S$ is subbase of $\mathcal{O}_V$, it suffices to show $(f+g)^{-1}(B)\in \mathcal{O}$ for all $B\in S$. So, let $B_r(v)\in S$. We have $$ x\in (f+g)^{-1}(B_r(v)) \iff \| f(x)+g(x)-v\| < r$$
But this is not really helpful, since $g(x)$ can basically be everything.
Let $U$ be open in $V$. Let $x_0 \in (f+g)^{-1}(U)$. Then $f(x_0)+g(x_0) \in U$, so there exists $r>0$ such that $B(f(x_0)+g(x_0),r) \subset U$. Consider $V=f^{-1}(B(f(x_0),r/2) \cap g^{-1}(B(f(x_0),r/2)$. This set is open in $X$ because it is an intersection of two open sets. Let $x\in V$. Now can you use triangle inequality for the norm to show that $x_0 \in V \subset (f+g)^{-1}(U)$?