A discrete cyclic subgroup of the general linear group

Computation. $A^k=\operatorname{diag}(2^k,2^{-k})$ for $k\in\mathbb{Z}$.

Uniformly discrete. Use the max-entry distance $d(X,Y)=\max_{i,j}|x_{ij}-y_{ij}|$ (any norm on $M_2$ is at least a constant times it; for the operator or Frobenius norm this distance is a lower bound). Let $j<k$.
- If $j\ge0$: $2^k-2^j\ge2^j(2-1)\ge1$.
- If $k\le0$: $2^{-j}-2^{-k}=2^{-k}(2^{k-j}-1)\ge2^{-k}\ge1$.
- If $j<0<k$: $2^k-2^j\ge2-\tfrac12=\tfrac32>1$.

So $d(A^j,A^k)\ge1$ for all $j\ne k$. In particular every point of $\Gamma$ is isolated: $\Gamma$ is discrete.

Closed. Suppose $A^{k_m}\to X$ in $M_2(\mathbb{R})$. A convergent sequence is Cauchy, so for large $m,m'$ we would have $d(A^{k_m},A^{k_{m'}})<1$, which forces $k_m=k_{m'}$. So the sequence is eventually constant and $X\in\Gamma$. Hence $\Gamma$ is closed in $M_2(\mathbb{R})$, and therefore in $GL_2(\mathbb{R})$.

Remark. Contrast with $\{R_{k\alpha}\}$ in the previous problem, which can be dense: powers of a matrix with eigenvalues of modulus $\ne1$ escape to infinity, while eigenvalues of modulus $1$ can make the powers accumulate. Lesson. Uniform discreteness gives closedness for free.