Here is a counterexample for surjectivity: $$B=k[x,y], A=B[X,Y]/(xX+yY-1)$$ It is easy to see that the image of the induced morphism is the plane minus the origin.
It is not hard neither to see that the extension is algebraically closed: If $a\in \text{Frac}(A)=k(x,y,X)$ is algebraic over $\text{Frac}(B)=k(x,y)$, then it must be in $k(x,y)$ as this is a purely transcendental field extension. As $k[x,y]$ is integrally closed, $a$ must actually be in $k[x,y]$.