It seems the following.
In general the answer is no, because compactness is different from sequential compactness. Let $\Bbb T=\{z\in\Bbb C:|z|=1\}$ be the unit circle endowed with the standard topology. Put $G={\Bbb T}^{\Bbb T}$. By Tychonov Theorem, $G$ is a compact space. Let $K=\{e\}$ be the trivial subgroup of $G$. Select an element $g=(g_z)_{z\in\Bbb T}\in G$ such that $g_z=z$ for each $z\in\Bbb T$. Suppose that there exists an increasing sequence $\{i_n\}$ of positive integers such that the sequence $\{g^{i_n}\}$ converges to the unit of the group $G$. Let $U_0=\{z\in\Bbb T: \operatorname{Re} z\ge 0\}$ be a neighborhood of the unit of the group $\Bbb T$. For each angle $z\in \Bbb T $ we can choose a number $n_z$ such that $i_nz\in U_0$ for each $n>n_z$. For each natural number $n$ put $G_n=\{z\in\Bbb T:n_z=n\}$. The continuity of power on the group $\Bbb T$ implies that the set $G_n$ is closed for each natural number $n$. Since $G=\bigcup_{n\in\Bbb N} G_n$, Baire Theorem implies that there exists a number $m$ such that a set $G_m$ has non-empty interior. Therefore there exists an open arc $U\subset G_m$ of the circle $\Bbb T$. Since the sequence $\{i_n\}$ is increasing, there exists a number $n$ such that $i_n>\max (m, 1/\mu(U))$, where $\mu$ is the standard measure on $\Bbb T$ such that $\mu(\Bbb T)=1$. But then $U_0\supset i_nG_m\supset i_n\overline U=\Bbb T$, a contradiction.