Proof, [Theorem 2.4, Zhu93]. (1): Since $G_{0}(A)$ is connected, it is open and closed. By Proposition 5.14.2, $G_{0}(A)$ coincides with the path component of $G(A)$ containing $1$.

Let $x, y \in G_{0}(A)$, then there exists paths $f, g \in C([0, 1]; G_{0}(A))$ such that $f(0) = g(0) = 1$, $f(1) = x$, and $g(1) = y$. The concatenation of $f$ and $xg$ then is a path from $1$ to $xy$, so $xy \in G_{0}(A)$. In addition, $t \mapsto f(t)^{-1}$ is a path from $1$ to $x^{-1}$, so $x^{-1}\in G_{0}(A)$ as well. Therefore $G_{0}(A)$ is a subgroup of $G(A)$.

Finally, let $x \in G(A)$, then $x^{-1}G_{0}(A)x$ is a connected subset of $G(A)$ containing $1$, so $x^{-1}G_{0}(A)x \subset G_{0}(A)$ and $G_{0}(A) \subset xG_{0}(A)x^{-1}$. Since the above holds for all $x \in G(A)$, $x^{-1}G_{0}(A)x = G_{0}(A)$.

(2): For each $x \in G(A)$, $xG_{0}(A)$ is connected, closed, and open, so it is a connected component by Lemma 5.12.5.$\square$