Necessary and sufficient condition for simplicial complex to be subspace

Question

Let $K$ be a simplicial complex and $|K|$ it's realization with the topology that $F\subset |K|$ is closed iff $f\cap \sigma$ is closed in $\sigma$, for every $\sigma\in K$. Then this topology coincides with the subspace topology if and only if for every $x\in |K|$, there exists a neighbourhood of $x$ that intersects finitely many simplexes in $K$.
Not exactly sure how to use the hypothesis. I'm thinking that this condition implies that every point has compact neighbourhood, thus it is locally compact. But then I need to omehow use that to show that if $F$ is such that $F\cap \sigma$ is closed, for every $\sigma$, then $F$ is closed in $R^n$, but how do I use compacteness to guarantee I can make F into a finite union?
Edit: I might have an answer, though it feels like I missed something: Let $F\subset |K|$ be such that, $F\cap\sigma$ is closed in $\sigma$, $\forall \sigma\in K$. This implies $F\cap \sigma = F_\sigma$, $F_\sigma$ closed in $\mathbb{R}^n$ for all $\sigma$. Take $x\in |K|\backslash F$. Let $U$ be neighbouhood of $x$ such that $U$ intercepts only $\sigma_1,...\sigma_n$. Then $x \in \cup(\sigma_i\backslash F_{\sigma_i})$=$(\cup\sigma_i)\backslash (\cup F_{\sigma_i})$ but that's open in $\cup\sigma_i$, so that there exists $U'$ open in $R^n$, $(\cup\sigma_i)\backslash (\cup F_{\sigma_i}) = U'\cap \cup \sigma_i$. But then $U\cap U'\subset \cup \sigma_i$ is a nieghbourhood of $x$ that's open in $|K|$ with the subspace topology, so that $F$ is closed, regarding to this topology too.
Is this ok?

Answer 1

Suppose there is an $x\in |K|$ such that every neighborhood intersects infinitely many simplexes. Choose neighborhoods $B(x,1/n)$, a sequence $(x_n)$ such that $x_n\neq x:n\in \mathbb N,\ x_n\to x$ and simplexes $\sigma_n$ such that $x_n\in B(x,1/n)\cap \sigma_n$. Then, $(x_n)$ is closed in $|K|$ for the finer topology. But $(x_n)$ is not closed for the subspace topology because $x$ is a limit point of $(x_n)$ but $x\notin (x_n)$.

On the other hand, given the hypothesis of the question, if $A$ is closed in $|K|$ for the finer topology, but not for the subspace topology, there is a sequence $(x_n)\subseteq A$ such that $x_n\to x\notin A$. Without loss of generality, assume that the $x_n\in \sigma_n$ such that $\sigma_i\neq \sigma_j$ if $i\neq j.$ Now, it's not hard to show that every neighborhood of $x$ intersects infinitely many simplexes and so we get a contradiction.