If $X$ is complete, then $A$ is complete if and only if $A$ is closed in $X$

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I'm doing Problem III.2.21 from textbook Analysis I by Amann.

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Could you please verify if my proof look fine or contains logical gaps/errors? Any suggestion is greatly appreciated. Thank you so much!

My attempt:

(i) Assume $(x_k)$ is a sequence in $A$ that converges to $x \in X$. Then $(x_k)$ is a Cauchy sequence in $X$, so $(x_k)$ is a Cauchy sequence in $A$. Because $A$ is complete, $x \in A$. Hence $A$ is closed in $X$.

(ii)

$\Longrightarrow$: This direction follows from (i).

$\Longleftarrow$: Assume $(x_k)$ is a Cauchy sequence in $A$. Then $(x_k)$ is a Cauchy sequence in $X$. Because $X$ is complete, $(x_k)$ converges to $x \in X$. Because $A$ is closed in $X$, $x \in A$. Hence $A$ is complete.