Why must a Turing Machine that Repeats the Same Configuration Twice be in an Infinite Loop?
This is because the transition function $\delta$ of a TM is deterministic. If the configuration is the same, then the arguments of $\delta$ are also the same, resulting in an infinite loop. Formally, this can be proved like @gnasher729 does.

Let's consider your example. If your $TM$ moves like the following,

$$
  (q_1, \dots00[0]111\dots) \xrightarrow{q_1, \text{right}} (q_1, \dots000[1]11\dots) \xrightarrow{q_1, \text{left}} (q_1, \dots00[0]111\dots)
$$

then the last step equals the first one, which means an infinite loop.