Metric Spaces - Distance Between sets and it's closures

[SqueeSpleen]

I was trying to prove:
$d(A,B) = d( \overline{A}, \overline{B} )$
I "proved" it using the following lemmas:
Lemma 1:
$d(A,B) = \inf \{ d(x,B) \}_{x \in A} = \inf \{ d(A,y) \}_{y \in B}$
(By definition we have: $d(A,B) = \inf \{ d(x,y) \}_{x \in A, y \in B}$ )
Lemma 2:
$d(x_{0},A) = d(x_{0}, \overline{A})$

Proof body:
$d(A,B) \underbrace{=}_{L1} \inf \{ d(x,B) \}_{x \in A} \underbrace{=}_{L2} \inf \{ d(x, \overline{B}) \}_{x \in A} \underbrace{=}_{L1} \inf \{ d(A,y) \}_{y \in \overline{B}} \underbrace{=}_{L2} \inf \{ d(\overline{A},y) \}_{y \in \overline{B}} \underbrace{=}_{L1} d(\overline{A}, \overline{B})$

The problem is that... I have the first lemma proved in the textbook, but the second lemma isn't in it and I couldn't prove it (That's why I said "proved" instead of proved).
I know that $d( x , \overline{B} ) \leq d(x,B)$ almost trivially because by definition is the infimum of a set and the infimum of a subset must be equal or higher.
But all the things I though are useful to prove $d( x , \overline{B} ) \leq d(x,B)$ not $d( x , \overline{B} ) \geq d(x,B)$.
If $d(x,B) = 0$ I know it's true because of a theorem that implies it.
But if $d(x,B) > 0$ I can't prove $d( x , \overline{B} ) \geq d(x,B)$

 

[micromass]

Assume by contradiction that $d(x,\overline{B})<d(x,B)$.

Take a sequence $(x_n)_n$ such that each $x_n\in \overline{B}$ and such that $d(x,x_n)\rightarrow d(x,\overline{B})$. Then there is some $n$ such that

$$d(x,x_n)<d(x,B)$$

Now try to find a contradiction. What is $d(x_n,B)$?

 

[SqueeSpleen]

If $x_{n} \in \overline{B}$ then for every $\varepsilon > 0$, in particular for every $\varepsilon \in (0,d(x,B)-d(x,x_{n}))$ there exists a $y \in B$ such that $d(x_{n},y) < \varepsilon$
Then $d(x,y) \leq d(x,x_{n})+d(x_{n},y) < d(x,B)$
But $y \in B$, so $d(x,B) \leq d(x,y)$ and we have $d(x,B) < d(x,B)$ which is a contradiction.

Thank you very much micromass.

 

[micromass]

Well done!