I’ve been trying to find a simplicial complex $K$, where $A$ and $B$ are subcomplexes, and $K=A\cup B$ to show that the Betti number
$\beta_i(K)\neq \beta_i(A)+\beta_i(B)-\beta_i(A\cap B)$
but every complex I come up with contradicts this for $i=0$, what would be a good example for this?
Here's an example (for $i=0$) where $\beta_0(K)\neq\beta_0(A)+\beta_0(B)-\beta_0(A\cap B)$. Let simplicial complex $K$ be the boundary of a triangle, with three vertices $[0]$, $[1]$, $[2]$, and three edges $[0,1]$, $[1,2]$, $[2,0]$. Let subcomplex $A$ be the union of two edges, containing all three vertices $[0]$, $[1]$, $[2]$, and only two edges $[0,1]$, $[1,2]$. Let subcomplex $B$ be the third remaining edge, containing vertices $[0]$, $[2]$ and only one edge $[2,0]$. We have $K=A\cup B$.
Note that $K$, $A$, and $B$ are each connected, giving $\beta_0(K)=\beta_0(A)=\beta_0(B)=1$. However, $A\cap B$ consists of two connected components (two vertices $[0]$ and $[2]$), giving $\beta_0(A\cap B)=2$. So we have $$\beta_0(K)=1\neq 0=1+1-2=\beta_0(A)+\beta_0(B)-\beta_0(A\cap B),$$ as desired. Note in this example that $\beta_1(K)=1$ is nonzero, agreeing with the Mayer-Vietoris long exact sequence (https://en.wikipedia.org/wiki/Mayer%E2%80%93Vietoris_sequence).