Prove that a set consisting of a sequence and its limit point is closed 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.
The proposition:
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...".
 A: As I said in my comment, your proof is ok. But here is the way I would have done it. I'll write the proof in a more formal way, because in math you can only talk loose after you master writing properly.
Let $y\in X\setminus A$. Let $\varepsilon=d(y,x)/2$. Then, by the convergence $x_n\to x$, there exists $n_0$ such that $x_n\in B_\varepsilon(x)$ for all $n\geq n_0$. So, for $n\geq n_0$, 
$$\tag{1}
d(x_n,y)> d(x,y)-d(x_n,x)>d(x,y)-\varepsilon=d(x,y)/2.
$$
Let $\delta=\min\{d(x,y),d(x_1,y),\ldots,d(x_{n_0},y)\}/2$. Then
$d(y,x_n)\geq\delta$ if $n\leq n_0$, and by ($1$) $d(y,x_n)\geq\delta$ if $n\geq n_0$. This shows that $B_\delta(y)$ has no intersection with $A$, i.e. $B_\delta(y)$ is contained in $X\setminus A$. As $y$ was arbitrary, this shows that $X\setminus A$ is open, i.e. $A$ is closed. 
A: What you really want to do is verify that $A$ satisfies your definition of closed set: that first of all any sequence $y_n \in A$ has a limit, i.e. that there exists $y$ so that $\lim y_n = y$, and second of all that $y \in A$. Informally speaking you want to say that every sequence in $A$ converges to a limit that is also in $A$ (note that the limit does not need to be $x$). This essentially eliminates the problems people pointed out in the comments. Based on this I bet you can write a new proof pretty quickly. Nevertheless, here is a quick outline of a solution:
If $y_n \in A$ you know that $y_n = x_{n_k}$ or $y_n = x$. From this you may conclude that either $y_n$ is eventually constant or $y_n$ converges to $x$. Either way you see that $y_n$ converges to a limit contained in $A$.
