I am looking at the proof of infinite set being equivalent to its proper subset.
In a part of proof 1, we have: $$S \cup (T \setminus S)= T$$ and $$S_{1} \cup (T \setminus S) = T \setminus S_{2}$$ I understand this part; however I have difficulty to see there is a bijection between $T$ and $T \setminus S_{2}$. It makes sense, but I am not sure why is that true.
You have established that there is a bijection $f$ from $S$ to $S_1$. Now, $T \smallsetminus S$ is disjoint with $S$ and hence with $S_1$ (since $S_1 \subseteq S$), i.e. $S \cap (T \smallsetminus S) = \emptyset = S_1 \cap (T \smallsetminus S)$. This means that the following function $h \colon S \cup (T \smallsetminus S) \to S_1 \cup (T \smallsetminus S)$ defined by: \begin{align} h(x) = \begin{cases} f(x) & \text{if } x \in S \\ x & \text{otherwise (i.e. if }x \in T \smallsetminus S\text{)} \end{cases} \end{align} is well-defined and is actually a bijection (since the "$f$ part" of $h$ is a bijection and the "identity part of $h$ is a bijection, and the two parts do not "overlap" and "cover" all the domain and codomain of $h$).
Finally, it turns out that $S \cup (T \smallsetminus S) = T$ and $S_1 \cup (T \smallsetminus S) = T \smallsetminus S_2$, thus $h$ is indeed a bijection from $T$ to $T \smallsetminus S_2$.