/-- Given a closed nonempty subset `s` of `Π (n : ℕ), E n`, there exists a Lipschitz retraction
onto this set, i.e., a Lipschitz map with range equal to `s`, equal to the identity on `s`. -/
theorem exists_lipschitz_retraction_of_isClosed {s : Set (∀ n, E n)} (hs : IsClosed s)
    (hne : s.Nonempty) :
    ∃ f : (∀ n, E n) → ∀ n, E n, (∀ x ∈ s, f x = x) ∧ range f = s ∧ LipschitzWith 1 f := by
  /- The map `f` is defined as follows. For `x ∈ s`, let `f x = x`. Otherwise, consider the longest
    prefix `w` that `x` shares with an element of `s`, and let `f x = z_w` where `z_w` is an element
    of `s` starting with `w`. All the desired properties are clear, except the fact that `f` is
    `1`-Lipschitz: if two points `x, y` belong to a common cylinder of length `n`, one should show
    that their images also belong to a common cylinder of length `n`. This is a case analysis:
    * if both `x, y ∈ s`, then this is clear.
    * if `x ∈ s` but `y ∉ s`, then the longest prefix `w` of `y` shared by an element of `s` is of
    length at least `n` (because of `x`), and then `f y` starts with `w` and therefore stays in the
    same length `n` cylinder.
    * if `x ∉ s`, `y ∉ s`, let `w` be the longest prefix of `x` shared by an element of `s`. If its
    length is `< n`, then it is also the longest prefix of `y`, and we get `f x = f y = z_w`.
    Otherwise, `f x` remains in the same `n`-cylinder as `x`. Similarly for `y`. Finally, `f x` and
    `f y` are again in the same `n`-cylinder, as desired. -/
  set f := fun x => if x ∈ s then x else (inter_cylinder_longestPrefix_nonempty hs hne x).some
  have fs : ∀ x ∈ s, f x = x := fun x xs => by simp [xs]
  refine' ⟨f, fs, _, _⟩
  -- check that the range of `f` is `s`.
  · apply Subset.antisymm
    · rintro x ⟨y, rfl⟩
      by_cases hy : y ∈ s
      · rwa [fs y hy]
      simpa [if_neg hy] using (inter_cylinder_longestPrefix_nonempty hs hne y).choose_spec.1
    · intro x hx
      rw [← fs x hx]
      exact mem_range_self _
  -- check that `f` is `1`-Lipschitz, by a case analysis.
  · refine LipschitzWith.mk_one fun x y => ?_
    -- exclude the trivial cases where `x = y`, or `f x = f y`.
    rcases eq_or_ne x y with (rfl | hxy)
    · simp
    rcases eq_or_ne (f x) (f y) with (h' | hfxfy)
    · simp [h', dist_nonneg]
    have I2 : cylinder x (firstDiff x y) = cylinder y (firstDiff x y) := by
      rw [← mem_cylinder_iff_eq]
      apply mem_cylinder_firstDiff
    suffices firstDiff x y ≤ firstDiff (f x) (f y) by
      simpa [dist_eq_of_ne hxy, dist_eq_of_ne hfxfy]
    -- case where `x ∈ s`
    by_cases xs : x ∈ s
    · rw [fs x xs] at hfxfy ⊢
      -- case where `y ∈ s`, trivial
      by_cases ys : y ∈ s
      · rw [fs y ys]
      -- case where `y ∉ s`
      have A : (s ∩ cylinder y (longestPrefix y s)).Nonempty :=
        inter_cylinder_longestPrefix_nonempty hs hne y
      have fy : f y = A.some := by simp_rw [if_neg ys]
      have I : cylinder A.some (firstDiff x y) = cylinder y (firstDiff x y) := by
        rw [← mem_cylinder_iff_eq, firstDiff_comm]
        apply cylinder_anti y _ A.some_mem.2
        exact firstDiff_le_longestPrefix hs ys xs
      rwa [← fy, ← I2, ← mem_cylinder_iff_eq, mem_cylinder_iff_le_firstDiff hfxfy.symm,
        firstDiff_comm _ x] at I
    -- case where `x ∉ s`
    · by_cases ys : y ∈ s
      -- case where `y ∈ s` (similar to the above)
      · have A : (s ∩ cylinder x (longestPrefix x s)).Nonempty :=
          inter_cylinder_longestPrefix_nonempty hs hne x
        have fx : f x = A.some := by simp_rw [if_neg xs]
        have I : cylinder A.some (firstDiff x y) = cylinder x (firstDiff x y) := by
          rw [← mem_cylinder_iff_eq]
          apply cylinder_anti x _ A.some_mem.2
          apply firstDiff_le_longestPrefix hs xs ys
        rw [fs y ys] at hfxfy ⊢
        rwa [← fx, I2, ← mem_cylinder_iff_eq, mem_cylinder_iff_le_firstDiff hfxfy] at I
      -- case where `y ∉ s`
      · have Ax : (s ∩ cylinder x (longestPrefix x s)).Nonempty :=
          inter_cylinder_longestPrefix_nonempty hs hne x
        have fx : f x = Ax.some := by simp_rw [if_neg xs]
        have Ay : (s ∩ cylinder y (longestPrefix y s)).Nonempty :=
          inter_cylinder_longestPrefix_nonempty hs hne y
        have fy : f y = Ay.some := by simp_rw [if_neg ys]
        -- case where the common prefix to `x` and `s`, or `y` and `s`, is shorter than the
        -- common part to `x` and `y` -- then `f x = f y`.
        by_cases H : longestPrefix x s < firstDiff x y ∨ longestPrefix y s < firstDiff x y
        · have : cylinder x (longestPrefix x s) = cylinder y (longestPrefix y s) := by
            cases' H with H H
            · 

Proof state:
case inl
E : ℕ → Type u_1
inst✝¹ : (n : ℕ) → TopologicalSpace (E n)
inst✝ : ∀ (n : ℕ), DiscreteTopology (E n)
s : Set ((n : ℕ) → E n)
hs : IsClosed s
hne : Set.Nonempty s
f : ((n : ℕ) → E n) → (n : ℕ) → E n :=
  fun x => if x ∈ s then x else Set.Nonempty.some (_ : Set.Nonempty (s ∩ cylinder x (longestPrefix x s)))
fs : ∀ x ∈ s, f x = x
x y : (n : ℕ) → E n
hxy : x ≠ y
hfxfy : f x ≠ f y
I2 : cylinder x (firstDiff x y) = cylinder y (firstDiff x y)
xs : x ∉ s
ys : y ∉ s
Ax : Set.Nonempty (s ∩ cylinder x (longestPrefix x s))
fx : f x = Set.Nonempty.some Ax
Ay : Set.Nonempty (s ∩ cylinder y (longestPrefix y s))
fy : f y = Set.Nonempty.some Ay
H : longestPrefix x s < firstDiff x y
⊢ cylinder x (longestPrefix x s) = cylinder y (longestPrefix y s)

Next tactic:
exact cylinder_longestPrefix_eq_of_longestPrefix_lt_firstDiff hs hne H xs ys