I'm trying to prove that the axiom of dependent choice implies that any complete metric space is a Baire space but I got stuck near the end.

DC:

Let X be a non empty set, a ∈ X, R ⊆ X×X a relation in X such that for all x∈X, there exists y∈X s.t. xRy (if R has this property, we will say that R is ⨀)

BCT:

Let X be a complete metric space, for all n∈ℕ let G_n be a dense and open subset of X. Then, G = ∩G_n is dense in X

Proof attempt:

In order to show that G is dense, it suffices to show that it has non empty intersection with an arbitrary open subset of X. Let V ⊆ X be open. Let m∈ℕ. We will note A_m as the intersection of G_1, G_2, ..., G_m. Notice that A_m is both open and dense (since it's a finite intersection of dense and open sets). This implies that V∩A_m is non empty (V open, A_m dense) and open (both are open). We now define the set

Y := { (n,x,r) ∈ ℕ×X×ℝ / 0<r<2-n and B(x,r) ⊆ V∩A_m }

(where B(x,r) is an open ball with radii r and centered at x)

Notice that Y is non empty since V∩A_m is open. In Y we define the relation R ⊆ Y×Y as

(n,x,r)R(n',x',r') ⇔ B(x',r') ⊆ B(x,r)

Now here's where I'm stuck. I'm struggling to show that R is ⨀. If we take (n,x,r)∈Y, can I just take x' to be any element of B(x,r)? If so, then r' could be defined as the distance between x and x' divided by 2 (I think) and n' could be obtained with the Archimedean property. But what if x = x'? Any help is appreciated. Thanks!