Metric Spaces - Distance Between sets and it's closures

I was trying to prove:

[itex]d(A,B) = d( \overline{A}, \overline{B} )[/itex]

I "proved" it using the following lemmas:

Lemma 1:

[itex]d(A,B) = \inf \{ d(x,B) \}_{x \in A} = \inf \{ d(A,y) \}_{y \in B}[/itex]

(By definition we have: [itex]d(A,B) = \inf \{ d(x,y) \}_{x \in A, y \in B}[/itex] )

Lemma 2:

[itex]d(x_{0},A) = d(x_{0}, \overline{A})[/itex]



Proof body:

[itex]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})[/itex]



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 [itex] d( x , \overline{B} ) \leq d(x,B) [/itex] 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 [itex] d( x , \overline{B} ) \leq d(x,B) [/itex] not [itex] d( x , \overline{B} ) \geq d(x,B) [/itex].

If [itex] d(x,B) = 0[/itex] I know it's true because of a theorem that implies it.

But if [itex] d(x,B) > 0[/itex] I can't prove [itex] d( x , \overline{B} ) \geq d(x,B) [/itex]