Recently I thought on one statement and I'm trying to prove it wrong or true, with the example in the latter
Suppose we have a normal-form game $G$ with two Nash equilibria in pure strategies. The question is if game $G_n$, $n$-times repeated game $G$, has exactly two subgame perfect equilibrium in pure stategies
No. You can get for combinatorial reasons more equilibria in pure strategies in which an equilibrium of the stage game gets played in every period. But there are also completely different equilibria:
Consider a simple coordination game in which both players always get the same payoff. There is a good $(G)$ and a bad $(B)$ action to coordinate on, and both players coordinating is better than not coordinating. $$ \begin{array}{c|cc} & G & B\\ \hline G& 4,4&0,0\\ B&0,0&1,1 \end{array} $$
We look at the three times repeated version of this game. We consider the strategy profile in which the players play on the equilibrium path $(G,B)$ in the first round and then $(G,G)$ in period $2$ and $3$. If someone deviates from this path, $(B,B)$ is played in the remaining periods. This is an SPNE in which strange behavior in the first period is made possible by the threat to switch to a worse continuation equilibrium.
There are versions of the folk-theorem for finitely repeated games that follow this logic.