Continuity of f implies Gf is closed. Assume that f is a continuous map. All elements of the set X × Y \ Gf have the form (x,f(x′)) where x≠x′ given the definition of Gf. Let N = U × V be a neighborhood of (x,f(x′)). If there is an x0 such that (x0,f(x0)) is contained within N then we can separate f(x0) and f(x′) with open sets according to the Hausdorff condition of Y , and replace V with V ′ where V ′ is the neighborhood containing f(x′). This yields a new neighorhood N′ = U × V ′ of (x,f(x′)) which doesn’t contain (x0,f(x0)). Doing this for all such (x0,f(x0)) in N will return to us a neighborhood of (x,f(x′)) which is disjoint from Gf. Thus X × Y \ Gf is open, and therefore Gf is closed.1
Gf being closed implies continuity of f. Assume that Gf is a closed subset of X ×Y . Let V be a neighborhood of f(x0) for some point x0 ∈ X. Then Y \V is a closed set yielding that B = Gf ∩ (X × (Y \V )) is also closed. Since Y is compact, then by Munkres exercise 26.7 π1(B) is closed as well. Now because B is the set of all points x × f(x) where f(x) ⁄∈ V , then U = X \ π1(B) is a neighborhood of x0 as f(x0) ∈ V and furthermore U ∈ f(X). Thus by Munkres Theorem 18.3 gives us that f is continuous.
Let {An} be a countable collection of closed subsets of a compact Hausdorff set X. Assume that each set in the collection has empty interior. For later contradiction assume that ∪An has nonempty interior, U. Let {xn} be a sequence of points where each xn is some point of An. We associate with this sequence, a sequence of subsets of U where V 1 is an open, nonempty subset of U such that x1 ⁄∈V 1 and each V n is an open, nonempty subset of V n-1 such that xn ⁄∈V n. We can find such a V n for each xn because of the foundation layed for us by step one of Munkres proof of Theorem 27.7 for a compact Hausdorff space such as X. Note that we need not concern ourselves with whether or not each xn is an isolated point as it is not contained in U since each An has no interior.
Now since each element of {V n} is nonempty and V n ⊂ V n-1 ⊂
⊂ V 1, then each element of {V n} is also nonempty
and V n ⊂V n-1 ⊂
⊂V 1 yielding that 
V n
is a collection of closed sets with the finite intersection property. Since our
space, X, is compact V = ∩V n is therefore nonempty.
However, since each xn is not in V , then by Munkres Lemma 26.4 we can find an open Un containing V with xn ⁄∈ Un. This implies that no limit point of V can be contained in any An, but since each An has no interior than no interior point of V can be contained within any An. Thus V has no interesection with ∪An, which contradicts the nonemptiness of V since V ⊂∪An. Therefore we must have that the interior of ∪An, U, is empty.
f is continuous. An isometry is continuous, easily seen by the plain, old ϵδ-definition of continuity on a metric space since for any ϵ we can set δ = ϵ and we will get that
since d(x,y) = d(f(x),f(y)).
f is injective If x and y are in X such that f(x) = f(y), then we have that d(f(x),f(y) = 0, but because f is an isometry, then this yields d(x,y) = 0 which can only be the case if x = y.
f is surjective For later contradiction, assume that f is not surjective. Then there is some a ∈ X with a ⁄∈ f(X). Then since X is Hausdorff and f(X) is compact (it’s the continuous image of X), then Munkres Lemma 26.4 allows us to find an ϵ such that the ϵ-neighborhood of a is disjoint from f(X). With this in mind, we construct a sequence recursively by defining xn to be f(xn-1) and with a base case of x1 = a. Hence for any points xi,xj of the sequence with i < j
by the definition of the sequence, and has
by the isometric property of f. However, the distance from a to any point of the image of f is greater than or
equal to ϵ, so d(xi,xj) ≥ ϵ, indicating that this particular sequence has no limit point. This contradicts the
compactnes of X as it would imply that X would not be limit-point compact. Hence f must be surjective.
Finally, we have that f is a continuous bijection from X to itself. This gives us that f-1 too is continuous, and so f is a homeomorphism.