Suppose you have a simplicial complex and a vertex $v$ which is not connected to any other vertex. Is $st(v)$ just the empty set? If you're looking at the inside of a simplex you don't look at anything on the outside edges and the vertex v is definitely lying on the outside here?
Similarly, is the link $lk(v)$ equal to $v$?
Recall that $st(v)$ is, by definition, the union of those open simplices of $K$ containing $v$. If you use the wikipedia definition, which drops the hipothesis of the simplices being open, the answer is the same.
If there is no other vertex connected to $v$, you are saying that $v$ is not a proper face of any other simplex $\tau$, or in other words, $\{v\}$ is a connected component of $K$ (in fact an isolated point). Therefore, $\{v\}$ is an open subset of $K$, and a 0-simplex. Hence, it is an open simplex containing $\{v\}.$ So, $\{v\} \subseteq st(v)$. But there are no more simplexes in $K$ containing $v$, so we conclude that $st(v)=\{v\}.$
On the other hand, $lk(v)$ is, by definition, $\overline{st(v)}-st(v)= \emptyset$, because $\{v\}$ is also a closed subspace of $K$.