Let $(X,d)$ be a metric space. Let $A$ be a subset of $X$.
Suppose I've found a sequence of elements $\ \{ x_{n} \}_{n=1}^{\infty} \subset \mathrm{cl}(A)$, such that $\lim\limits_{n \to \infty}\{x_{n}\} = x$, where $x \in X$.
How do I formally prove that $x \in \mathrm{cl}(A)$?
My thought was to define a set $B := \mathrm{cl}(A)$. Since $\ \{ x_{n} \}_{n=1}^{\infty} \subset B$ converges to some $x \in X$, it follows by the definition of a closure that $x \in \mathrm{cl}(B)$. So if I can show that $\mathrm{cl}(B) = B$, then I am done.
Is there an easier way?
For simplicity, let me replace $\{x_n\}$ with a subsequence (that I still name $\{x_n\}$) such that $d(x_n,x)<1/n$.
For each $n$, choose $y_n\in A$ such that $d(x_n,y_n)<1/n$ (such $y_n$ exists because $x_n$ is in the closure of $A$). Then $$ d(x,y_n)\leq d(x,x_n)+d(x_n,y_n)<\frac2n. $$ This shows that we can find elements in $A$ arbitrarily close to $x$, and so $x$ is in the closure of $A$.