Can someone please check whether the following simple proof is "mathematical"? Is it correct, complete, rigid? Can it be simplified? I'm a complete autodidact so I'm looking for someone to give me feedback to gain experience in writing proofs... This is also my first question on MSE.
Let $(X, d)$ be a metric space and $x_n \to x$ where each $x_n \in X$ and $x \in X$. Let $A$ be the subset of $X$ which consists of $x$ and all of the points $x_n$. Prove that $A$ is closed in $(X, d)$.
My tentative to prove this:
We first show that all infinite sequences in $A$ converge to $x$: Let $y \in X$, $y \ne x$. Then there is some open ball $B_\epsilon(x)$ with $\epsilon < d(x,y)$ containing all but finitely many elements of $A$. As $y \notin B_\epsilon(x)$ there can be no infinite sequence in $A$ converging to $y$. Consequently all infinite series in $A$ converge to a point in $A$ which therefore must be a closed set.
Edited: As rightly pointed out in the comments, I should have written in the first sentence "...sequences with infinitely many distinct terms and which converge to some point of $X$" and the last sentence should be "Consequently all infinite sequences...".