Denote with T𝛼 the topology of 𝑋𝛼.
Consider the subbasis A𝑋 ={𝜋−1𝛼𝑈 :𝑈 ∈T𝛼 :𝛼 ∈𝐴} of 𝑋,
and let 𝑈 ∈A𝑋 be a subbasic open set.
Then 𝑈 =𝜋−1𝛼(𝑉) for some 𝑉 ∈T𝛼,
and since 𝜋𝛼 is surjective, 𝜋𝛼(𝜋−1𝛼(𝑉)) =𝑉 =𝜋𝛼(𝑈).
Hence 𝜋𝛼(𝑈) is open.
Thus, by Proving open map with a subbasis, each 𝜋𝛼 is an open map.