Topology Proof: g:(X1 x {p2})→ X1 is a homeomorphism

I've outlined the proof but could use some help with the details in the simpler sections (eg: proving a bijection)

For any p₂ ∈ X₂
π₁|_(x1x{p2}): (X₁ x {p₂})→ X₁ is a homeomorphism (ie: bijective, continuous and open)

Proof: Consider any p₂ ∈ X₂
Let g=π₁|_(x1x{p2})
Claim 1: g: (X₁ x {p₂})→ X₁ is a bijection
Claim 1.1: If g(p)=g(q), then p=q (ie: g is injective or one-to-one)
Proof: Let g(p)=g(q)
[I know this is simple, but I haven’t done this sort of proof in a while…how do I show p=q? Can I say g(g(p))=g(g(q))? Then p=q?]
Claim 1.2: For every x∈ X₁, there exists p∈ (X₁ x {p₂}) such that g(p)=x (ie: g is surjective or onto)
Proof: consider any x∈ X₁
[Again, this seems like a simple proof, but I could use some help here]
Therefore, g: (X₁ x {p₂})→ X₁ is injective from Claim 1.1
And g: (X₁ x {p₂})→ X₁ is surjective from Claim 1.2
Therefore, g: (X₁ x {p₂})→ X₁ is a bijection

Claim 2: g: (X₁ x {p₂})→ X₁ is continuous
Proof: We know f₁: (X₁ x X₂) → X₁ is a continuous open surjection
Therefore, g: (X₁ x {p₂})→ X₁ is continuous [Can we say this follows directly because p₂ ∈ X₂ or is there a way to add more detail here?]

Claim 3: If B={(G₁x{p₂})| G₁∈Τ₁} is a base for Τ_{X1 x X2}|_{X1 x {p2}} then f₁[(G₁ x {p₂})]∈ Τ_{X1 x X2}|_X1
Proof: Let B be a base for Τ_{X1 x X2}|_{X1 x {p2}}
Then f₁^-1[(G₁ x {p₂})]∈ Τ_{X1 x X2}|_{X1 x {p2}}
Then f₁[(G₁ x {p₂})]∈ Τ_{X1 x X2}|_X1
Ie: g: (X₁ x {p₂})→ X₁ is open
So g: (X₁ x {p₂})→ X₁ is a homeomorphism from claims 1,2 and 3