The question:
I am required to show the following: Let $K$ be a finite simplicial complex, let $x \in |K|$ and suppose that $U$ is an open set containing $x$. Then there is an open connected set $V$ such that $x \in V \subseteq U$.
My attempt:
Let $D$ be the disjoint union of simplices from which $|K|$ is obtained, and let $q:D \to |K|$ be the quotient map.
$q^{-1}(U)$ is open in $D$, by definition of quotient spaces.
We take it as standard that the restriction of $q$ to the insides of the simplices in $D$ is a bijection, so there is exactly one simplex $\sigma$ in $D$ such that $x \in q($inside$(\sigma))$.
Then there is some $\overline{x}$ in $\sigma$ such that $q(\overline{x}) = x$.
$\overline{x} \in q^{-1}(U) \cap ($inside$(\sigma))$, which is an open subset of inside$(\sigma)$, and hence is an open subset of $\mathbb{R}^n$ for some $n$. Thus, there exists $\epsilon > 0$ such that $B(\overline{x},\epsilon) \subseteq q^{-1}(U) \cap ($inside$(\sigma))$.
$B(\overline{x}, \epsilon)$ is connected, open in $D$, and contains $\overline{x}$.
Thus, $q(B(\overline{x}, \epsilon))$ is connected (by the continuity of $q$), open in $|K|$ (by definition of a quotient space), and contains $x$. So it is the desired $V$.
Concerns:
I feel as if I'm overcomplicating this hugely - I struggle to know what I'm "allowed" to assume as intuitively clear in this topic.
I'm also not sure that the argument I'm using even works, as it relies on jumping around a lot of different topologies and it's possible that I'm making unwarranted assumptions.