Initially, I know that:
Theorem. All Cauchy sequence are bounded.
My proof trying. Let $x_n$ be a Cauchy. Then, $x_n$ is converget by the $2$. So, by the $1$, $x_n$ is bounded. Hence, directlty, did I prove the theorem? Because $x_n$ is bounded.
Yes, that is correct. It's instructive to also prove it directly using the definition. (Plus it will be useful later on when you encounter metric spaces that aren't complete.)
Suppose $(x_n)_n$ is a Cauchy sequence. Then there exists $N\in\mathbb{N}$ such that $|x_n-x_m|<1$ for all $n,m\geq N$. In particular, we have $|x_n-x_N|<1$ for all $n\geq N$. Thus $|x_n|\leq |x_n-x_N|+|x_N| < 1+|x_N|$ for all $n\geq N$, so that if $M:=\max\{|x_1|,\ldots,|x_{N-1}|,1+|x_N|\}$, then we have $|x_n|\leq M$ for all $n\in\mathbb{N}$.
I think a better proof would be a direct proof (since it would show your understanding of the material). Let $x_n$ be a Cauchy sequence. Since $x_n$ is Cauchy, we know that $\forall \epsilon$ $\exists N_{\epsilon}$ such that $n,m > N$ then $|x_n - x_m| < \epsilon$. Now, fix $\epsilon$ as some real number and consider $\{B_{\epsilon}(x_i): i \leq N_{\epsilon} +1\}$. Now, for each $x_j$ in our sequence, we have that $x_j \in \bigcup_{i=1}^{N_\epsilon +1}B_{\epsilon}(x_i)$. Now, if you show that $\bigcup_{i=1}^{N_\epsilon +1}B_{\epsilon}(x_i)$ is bounded, you will show that your Cauchy sequence is bounded.
Your proof is correct. In a complete metric space, the set of all Cauchy sequences is equal to the set of all convergent sequences.
The more general point, which you seem to be unsure about, is that if two sets $X$ and $Y$ are equal, then any proposition concerning one is also true for the other. In other words, for any predicate $P$, $$X=Y \land \forall x \in X, P(x) \Longrightarrow \forall y \in Y, P(y)$$
