Is this a valid proof that if an element $x$, is within the closure of a set $A$ (in a metric space) then $\exists (a_n) \in A : a_n \rightarrow x$ First form a sequence of epsilons $(\epsilon_n): \epsilon_1> \epsilon_2>\epsilon_3>\cdots$
    if $x \in \bar{A}$ then by the definition of the closure we have $$\forall \epsilon>0 , \exists a\in A :  d(x,a)< \epsilon$$
    Now for each $\epsilon_n$ there exists an $a_n$ such that $d(x,a_n)< \epsilon_n$ hence $$d(x,a_n)< \epsilon_n <\epsilon_{n-1}<\cdots< \epsilon_1 $$
    as constructed.
So we have $$\forall \epsilon_1>0 , \exists N \in \text{the natural numbers}:  d(x,a_n)< \epsilon_1,  \forall n>N$$
and so $a_n \rightarrow x$ as $n \rightarrow \infty$
(N.B Since this list can be constructed regardless of the magnitude of $\epsilon_1$ we really are free to choose any $\epsilon_1$ we like, as long as it is positive.)
EDIT: Thanks for your answers, I'd up vote you all but I don't have enough reputation
 A: It's not enough to say $\varepsilon_1 > \varepsilon_2>\varepsilon_3 > \cdots.$ You also need to say $\lim\limits_{n\,\to\,\infty} \varepsilon_n =0.$
For example, if $\varepsilon_1 = 1.1$ and $\varepsilon_2=1.01$ and $\varepsilon_3 = 1.001$ and so on, then you do have $\varepsilon_1 > \varepsilon_2>\varepsilon_3 > \cdots$ but you do not have $\lim\limits_{n\,\to\,\infty} \varepsilon_n =0,$ so that will not serve.
I would start with $\varepsilon_n = \dfrac 1 n$ or some other specific sequence whose terms decrease and approach $0,$ and after that proceed as you did.
A: Your idea is what is needed. However, imagine that you take $\varepsilon_n=1+\dfrac{1}{n}$. We still have $\varepsilon_1>\varepsilon_2>\dots$, but nothing ensures that $d(x,a_n)\to 0$. So you need to ensure that $\varepsilon_n\to 0$ as $n\to\infty$. A classical choice is $\varepsilon_n=\dfrac{1}{n}$.
In the proof you don't need to go back to $\varepsilon_1$. You have $$ \forall n\in\mathbb{N},\,d(x,a_n)\le\varepsilon_n.$$
Hence, as $\varepsilon_n\to 0$, you can conclude that $x=\lim_{n\to +\infty}a_n$ as desired.
A final remark: this is generally not mentioned in proofs, but when you have
$$\forall n\in\mathbb{N},\exists a_n\in A,d(x,a)\le\varepsilon_n$$
still this is not rigorous enough to justify the existence of a sequence $(a_n)_{n\in\mathbb{N}}$. The justification is the axiom of choice.
Edit
The existence of such a sequence can be justified using the axiom of choice. There are many different equivalent forms of this axiom. The one that I find most "natural" for our purpose is the following:

Let $\mathcal{C}$ be a nonempty collection of nonempty sets. Then there exists a function $f$ (called a choice function) such that
$$\forall S\in\mathcal{C},f(S)\in S.$$

In other words, the image by $f$ of a set $S\in\mathcal{C}$ is an element "chosen" from $S$.
We can now justify the existence of your sequence. Remember that a sequence $(a_n)_{n\in\mathbb{N}}$ of elements of $X$ is a function $a:\mathbb{N}\to X$, and writing $a_n$ instead of $a(n)$. The key is:
$$\forall n\in\mathbb{N},\exists a\in A,d(x,a)\le\varepsilon_n.$$
So, for $n\in\mathbb{N}$, let $S_n=\{a\in A\mid d(x,a)\le\varepsilon_n\}$, which is nonempty by what's above, and let $\mathcal{C}=\{S_n\mid n\in\mathbb{N}\}$ and $f:\mathcal{C}\to X$ be the corresponding choice function. Then the sequence $a_n=f(S_n)$ satisfies
$$\forall n\in\mathbb{N},d(x,a_n)\le\varepsilon_n.$$
Remark: This kind of construction of sequences oftently used in proofs in textbooks without even mentioning the axiom of choice. So I guess you shouldn't feel obliged to always mention it, but it's important to know how it's used.
A: Yes. Given $\epsilon_1>0,$ the problem is how to construct the sequence $(\epsilon_n) $ such that
$$\epsilon_1>\epsilon_2>\cdots>\epsilon_n $$
You can take for example
$$\epsilon_n=\frac {\epsilon_1}{n} $$
then
$$d (x,a_n)<\frac {\epsilon_1}{n} $$
and squeeze.
