If $(X, d)$ is a complete metric system and $Y \subset X$ is closed, then $(Y, d)$ is a complete metric system

Question

I begin with an arbitrary Cauchy sequence $\{ a_n \}$ in $Y$. As $Y \subset X$, we have that $\{a_n \} $is a Cauchy sequence in $X$. As $X$ is a complete metric space, $\{a_n \}$ converges to some $a \in X$. At this point, I want to show that $a$ is a limit point of $Y$ by using points from the definition of convergence. For instance, as $\{a_n\} \rightarrow a$, fix $\epsilon >0$ and there is $N_\epsilon \in \mathbb{N}$ such that $n \geq N_\epsilon$ implies that $d(a_n, a) < \epsilon$. But my question is how can we be sure that our sequence $\{a_n \}$ is not just $a_n = a$ for all $n \in \mathbb{N}$? This sequence is Cauchy, but doesn't grant any distinct points to be able to show that $a$ is a limit point of $Y$ directly. Once I have this, I know that $a \in Y$ as $Y$ is closed, and then $Y$ is complete.

Answer 1

If $a_n = a$ for all $n$ (or just one), then $a\in Y$ as $a_n\in Y$ for all $n$ by assumption.

Answer 2

the old way i remember doing this is this:

let X be complete, with $Y \subset_{closed} X$, let $x_n$ be a cauchy sequence in Y, since X is complete $x_n \longrightarrow x$ for some $x \in X$, since Y is closed, $x \in Y$

so long as x is a limit point for some convergent sequence in Y, and Y is closed x must be contained in Y, no matter what form it takes.