Proposition.
If Y is compact, the projection map π₁ : X × Y → X is a closed map.
Proof
Let A be closed in X × Y.
W := X - π₁(A)
Fix an x ∈ W
∀y ∈ Y, x × y ∈ X × Y - A and X × Y - A is open.
⇒ ∃open Vy ⊂ X, Uy ⊂ Y s.t. x × y ∈ Vy × Uy ⊂ X × Y - A ( x ∈ Vy and y ∈ Uy )
⇒ { Vy × Uy | y ∈ Y } is an open cover of x × Y contained in X × Y - A
Y is compact, so x × Y is compact.
⇒ ∃ finite subcover { Vi × Ui | i = 1, 2, ..., n } of x × Y contained in X × Y - A
⇒ Vi × Ui and A are disjoint for all i
⇒ ∀x' ∈ π₁(A), ∃Vi s.t. x'∉Vi
⇒ ∩Vi is open, and disjoint with π₁(A) (therefore contained in W)
[alternative method]
starting from
∃ finite subcover { Vi × Ui | i = 1, 2, ..., n } of x × Y contained in X × Y - A
⇒ { Vi × Ui | i = 1, 2, ..., n } covers the open set ∩Vi × Y and is disjoint with A
⇒ ∩Vi × Y is open and disjoint with A (Here, ∩Vi × Y is a TUBE containing the slice x × Y and contained in X × Y - A : refer to Tube Lemma)
⇒ π₁(∩Vi × Y) = ∩Vi is open and disjoint with π₁(A) (therefore contained in W)
This concludes that for all x ∈ W, there exists an open neighborhood ∩Vi contained in W.
⇒ W is open
⇒ π₁(A) is closed.
∴ π₁ is a closed map.
End of proof.
* Note. What if Y is not compact?
Let X = Y = ℝ in the standard topology
{ x × y | xy = 1 } is closed.
But its projection, which is ℝ - {0} is not closed.
So, the projection is not a closed map.