Let $K$ be a 2-dimensional simplicial complex embedded in $\mathbb{R}^3$. The link of a vertex $v$ in $K$ is the graph $\text{Lk}(v)$ whose vertices are the edges incident to $v$ in $K$. Two vertices in $\text{Lk}(v)$ are adjacent if and only if their corresponding edges in $K$ are incident to a common triangle. For example, if $K$ is a manifold then $\text{Lk}(v)$ is a simple cycle for all $v$.
Is it true that if $K$ is embedded in $\mathbb{R}^3$ then $\text{Lk}(v)$ is a planar graph for all $v$?
Yes, this is true, at least if you add one assumption, namely that for each simplex $\sigma$ of $K$ with vertex set $v_0,...,v_K$, $\sigma$ is equal to the ordinary simplex of $\mathbb R^3$ spanned by those points, meaning $$\sigma = \{t_0 v_1 + ... + t_K v_K \mid t_0+...+t_K=1, \,\text{and $t_k \ge 0$ for $0 \le k \le K$}\} $$ Under that assumption, for each vertex $v \in K$ and for each sufficiently small $r > 0$, if we let $S(v,r)$ be the sphere of radius $r$ centered on $v$ then $K \cap S(v,r)$ is an embedding of $\text{Lk}(v)$ into the sphere $S(v,r)$. And, of course, a graph is embeddeable into a sphere if and only if it is embeddable into the plane.