Suppose that $K_1 \supset K_2 \supset K_3 \supset \dots $ is a nested sequence of compact connected subsets of $S^2$ such that $\pi_1(K_j)\simeq \mathbb{Z}$ for all $j$. Prove or provide a counterexample for each of the following statements:
a. $\bigcap_{j=1}^\infty K_j$ is connected.
b. $\pi_1(\bigcap_{j=1}^\infty K_j)\simeq \mathbb{Z}$.
a. is true.
suppose, $K_\infty:=\bigcap K_i$ is not connected; then there exist two open sets $U_1,U_2$ such that $U_1\cap U_2=\emptyset$, $K_\infty\subset U_1\cup U_2$ and $U_j\cap K_\infty\ne\emptyset$.
then $L_i:=K_i\cap(S^2\setminus U_1\cup U_2)\ne\emptyset$, because $K_i\cap U_j$ is nonempty and $K_i$ are connected; also, $L_i$ are closed in $S^2$. so we have a contradiction: $\emptyset\ne\bigcap L_i\cap(U_1\cup U_2)$.
b. is false.
just take a collection of discs $\{D_i\}$ with radius of $D_i$ equals to $1/2^{i}$, such that $D_{i+1}\subset D_i$, and then let $K_i=D_i\setminus(\text{small open disc, disjoint with }D_{i+1})$.
$\pi_1(K_\infty)=0$ because all the maps $\pi_1(K_i)\to \pi_1(K_{i-1})$ are zero, so the limit of the diagram will be zero too. the picture like that:
PS: if in a. you need to path connectness, of course the answer will be false too, find an example yourself