Let $d$ and $p$ be two metrics on $X$ such that $$\frac{1}{2022}p(x,y) \le d(x,y) \le \frac{p(x,y)}{p(x,y)+1},$$ for all $x,y \in X$. Show that if $(X,p)$ is complete, then so is $(X,d)$.
Attempt: Let $(x_n)$ be an arbitrary Cauchy sequence in $(X,d)$. We'll show that $(x_n)$ converges in $(X,d)$, say, to $x \in X$. Let $\varepsilon>0$ be arbitrarily given. Since $(X,p)$ is complete, there exists $N \in \Bbb N$ such that for any $n \in \Bbb N$ with $n \ge N$, we have $p(x_n,x)<\varepsilon.$ Notice that since $p$ is a metric, then $p \ge 0$, so that $p+1 \ge 1$, which means $\frac{1}{p+1} \le 1$.
Hence, for any $\varepsilon>0$, there exists $N \in \Bbb N$ such that for any $n \in \Bbb N$ with $n \ge N$, we have \begin{align*} d(x_n,x) &\le \frac{p(x_n,x)}{p(x_n,x)+1} \\ &\le \frac{p(x_n,x)}{1} \\ &= p(x_n,x) \\ &< \varepsilon. \end{align*} Thus, $x_n \to x$ in $(X,d)$. Therefore, $(X,d)$ is complete, as desired.
Is the above approach correct?
Your approach is fine, but you don't define $x$ clearly. You should say "there exist $n \in \mathbb{N}$ and $x \in X$ such that" where you define $n$, and you should either delete the sentence starting "We'll show that", or change the sentence to finish "converges to a point in $X$" or "converges, with a limit $x \in X$".
Before you use the fact that $(X,p)$ is complete, you can't define the limit of $(x_n)$ - you can only say "Let $x \in X$ denote the limit point of $(x_n)$, if it exists". Any sentence starting "We'll show that" is a signposting sentence, and that means that this sentence can contain things that you'll prove later - in particular, you can say "We'll show that $(x_n)$ has a limit $x \in X$" instead of "We'll show that $(x_n)$ has a limit, which we'll call $x \in X$ if it exists". But the rest of your proof can't depend on that signposting sentence, because the signposting sentence depends on the proof.