Closure of a set in a metrizable space is $\{y \in X : \rho(E, y) = 0\}$ Let $X$ be a metrizable space whose topology is generated by a metric $\rho$. 
Prove: The closure of a set $E \subseteq X$ is given by $\tilde E = \{y \in X : \rho(E, y) = 0\}$.
Proof (So far):
The definition for closure I am given is $\overline E = \bigcap \{K$ closed $ : E \subseteq K\}$. I want to show that the above definition $\tilde E = \{y \in X : \rho(E, x) = \inf_{e \in E} \rho(e, x) = 0\}$ is equal to $\overline E$.
To show that $\overline E \subseteq \tilde E$ I considered $X \setminus \tilde E$. Let $y \in X \setminus \tilde E$, then $\rho(E, y) = \varepsilon > 0$ (otherwise it would be in $\tilde E$ a contradiction). Consider $U_\rho(x, \varepsilon/2) = \{y \in X : \rho(x, y) < \varepsilon/2\} \subseteq X \setminus \tilde E$. Conclude that $X \setminus \tilde E$ is open and thus $\tilde E$ is closed. But $E \subseteq \tilde E$ since $\rho(E, E) = 0$. By definition of $\overline E$, we can conclude that $\overline E \subseteq \tilde E$.
I want to show $\tilde E \subseteq \overline E$. Let $x \in \tilde E$, then by defintition $\rho(E, x) = \inf_{e \in E} \rho(e, x) = 0$. Suppose by way of contradiction that $x \notin \overline E$, then there exists a closed set $K$ with $E \subseteq K$ such that $x \in X \setminus K$. I want to contradict that $\rho(E, x) = 0$ by showing it's grater than zero but have not been able to do this.
 A: To continue where you left off, we want to show that $\rho (E,x) > 0$. Since $\rho (K,x) \leq \rho (E,x)$, it suffices to prove that $\rho (K,x) > 0$. Since $K$ is closed, $\exists \epsilon > 0$ such that $B(x,\epsilon) \subset X\setminus K$, and so, for any $y \in K$, $d(x,y) > \epsilon$. Hence, $\rho (K,x) \geq \epsilon > 0$.
Hence, we have shown that if $x\notin \overline{E}$, then $\rho (E,x) > 0$, which proves what you want.
A: You can do the following. Suppose $x\in\tilde E$, that is $d(x,E)=0$. By definition of the infimum then, for each $\varepsilon >0$ there exists $y\in E$ such that $d(x,y)<\varepsilon$. Pick $\varepsilon=1,1/2,\ldots$ to obtain a sequence of points $x_n\in E$ with $d(x,x_n)<n^{-1}$. It follows that $x_n\to x$, so $x\in \overline E \blacktriangleleft$
The above is justified by the following:

PROP Define $\overline E=\bigcap\{F:F\text{ closed and } E\subseteq F\}$. Then $x\in\overline E$ if every nbhd $U_x$ of $x$ meets $E$.

PROOF Suppose that there is some $U_x$ such that $E\cap U_x=\varnothing$. Then $X\smallsetminus U_x$ is closed and contains $x$, so $E\subsetneq X\smallsetminus U_x$ and $x\notin \overline E$. Now suppose $x\notin \overline E$. Then there is some closed set $F_x$ containing $E$ that doesn't contain $x$. It follows $U_x=X\smallsetminus F_x$ is an open set containing $x$ that doesn't meet $E\blacktriangleleft$. 
DEF Let $(X,\tau)$ be a topological space. Let $x_1,x_2,\ldots$ be a sequence in $X$. We say $x_n\to x$ if for every nbhd $U$ of $x$ there is some $N$ such that $n\geqslant N$ implies $x_n\in U$. In particular (you can show that) if $(X,\rho)$ is a metric space, $x_n\to x\iff \rho(x_n,x)\to 0$ in $\Bbb R$.

THM Let $(X,\rho)$ be a metric space (more generally, a first countable space), $E\subseteq X$. Then $x\in \overline E$ iff there is a sequence of points in $E$ that converges to $x$.

P Suppose that $x\in\overline E$. Pick a countable nested nbhd base of $x$, $U_1\supseteq U_2\supseteq U_2\supseteq\cdots$. By the proposition above, there is some $x_n\in U_n\cap E$. By definition of nbhd base, $x_n\to x$. Conversely, suppose we have a sequence $x_n\in E$ converging to $x$. Then every nbhd $U$ of $x$ contains a tail of $x_1,x_2,\ldots$; so it meets $E$, hence $x\in \overline E$ by the proposition $\blacktriangleleft$.
ADD Note that using the above, we can write any closed set in a metric space as $$E=\bigcap_{n\geqslant 1} \{x:d(x,E)<n^{-1}\}$$
i.e. every closed set in a metric space is a $G_\delta$, thus every open set is an $F_\sigma$.
