Let $\mathbb T_k$ be a d-dimensional torus with sides of length $k$, and let $S_n^{(k)}$ denote the lazy simple random walk on $\mathbb T_k$ (which jumps to each of its neighbors with equal probability, and does not jump with probability $\tfrac{1}{2}$), starting at the origin. Denote \begin{equation} p_n^{(k)}(x) = \mathbb P\{S_n^{(k)} = x\} \quad \text{for all} \quad x \in \mathbb T_k, \quad k, n \in \mathbb N, \end{equation}Let $(t_k)$ be a sequence satisfying $t_k / k^2 \to 0$ as $k \to \infty$. Let $\mu^{(k)}$ denote the uniform distribution on $\mathbb T_k$.
Prove that \begin{equation} \|p_{t_k}^{(k)} - \mu^{(k)}\|_\text{TV} \xrightarrow{k \to \infty} 1. \end{equation}
Apart from the fact that $\mu^{(k)}$ is also the stationary distribution of the Markov Chain corresponding to the random walk, and that $p_{t_k}^{(k)}$ tends to be small for values far from the origin, I have literally no idea where to start, so any help is much appreciated!
EDIT: As mentioned in the comments, 'neighbors' as in the problem description deserves some clarification. The set of neighbors of a vertex here means the set of adjacent vertices viewed on the lattice in $(\mathbb Z/ k \mathbb Z)^d \cong \mathbb T_k$.
I think application of Sangchul's idea to your problem is easier than your comment indicates. To show $\|P - Q\|_{\mathrm{TV}}$ is close to $1$ it suffices to exhibit an event $A$ for which $P(A)$ is close to $1$ and $Q(A)$ is close to $0$. (Where $P$ is the law of $S^{(k)}_n$, and $Q$ is the uniform distribution.) In your case (unless I've made an algebra mistake somewhere), $A$ can be as simple as the first component of $S^{(k)}_n$ obeying $|S^{(k)}_n|\le \sqrt n \lambda$ where $\lambda$ will chosen later.
The first component of $S^{(k)}_n$ is a 1-d random walk with step distribution concentrated at $\pm1$ with probabilities $1/4d$ each, and on $0$ with prob. $1-1/2d$. So its variance is $1/2d$. $P$ concentrates its mass near $0$, and $Q$ spreads it out. (There is a slight sloppiness here: the first component of $S^{(k)}_n$ is the mod $k$ reduction of an integer random variable with variance $n/2d$, and hence concentrated near the origin; modular reduction makes first component of $S^{(k)}_n$ even more concentrated near the mod $k$ origin.)
Then Chebyshev gives $P(A)\ge 1-1/(2d\lambda^2)$, and obviously $Q(A)\le(1+ 2\sqrt n\lambda)/k$. You want $n=(\epsilon_k k)^2$, where $\epsilon_k\to0$, so choice of $\lambda=1/\epsilon_n$ makes $P(A)\to 1$ and $Q(A)\to 0$.