Strict inequalities while solving uniform continuity problems in real analysis

Problem - Is g(x) = √ (x² + 1) uniformly continuous on (0, 1) ?

Solution given -

|g(x)−g(y)| = ​|√ (x²+1)​−√ (y²+1)​​| = |x² + y²|/(√ (x²+1)​+√(y²+1))

Since x,y∈(0,1), we know ∣x+y∣≤1+1=2

Also, √(x²+1)≥√(0²+1)=1, so:

√(x²+1)+√(y²+1)≥1+1=2.

Using the bounds for |x+y| and √(x²+1)+√(y²+1):

∣g(x)−g(y)∣≤∣x−y∣⋅2/2=|x−y|.

so we can use  δ=ϵ

My Problem :

Since (0,1) is an open interval then 0,1 ∉ (0,1) and hence |x+y| < 2 and not |x+y| ≤ 2.

And the same goes for √(x²+1)+√(y²+1)≥1+1=2. Instead it should be √(x²+1)+√(y²+1)>1+1=2.

Which further gives ∣g(x)−g(y)∣<∣x−y∣⋅2/2=|x−y|.

Am I correct or incorrect please tell me if I am right or wrong this stuff is really confusing please help me.