Diameter of a subset of an open ball

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Let A be a subset of a metric space such that A ⊆ B (p, r) for some p ∈ X and r > 0.
Show that diam(A) ≤ 2r.
B(p,r)=(p-r,p+r)
diam( B(p,r) )=sup{d(a,b)│a,b∈B(p,r) }=d(p-r,p+r)= 2r
Since A ⊆ B (p, r), the diameter of A is less than the diameter of B (p, r):
diam(A)≤2r

Is it true and enough? I think I have missed something in my argument.
 
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p is a point in your space, r is a real number. You can't add them. That makes no sense at all. Try using the triangle inequality.
 
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