Is there a way to avoid uniform convergence in Urysohn's extension theorem? In the remarkable book Rings of Continuous Functions, by Gillman and Jerison, I came across with the following theorem:
Urysohn's Extension Theorem (UT). A subspace $S$ of $X$ is $C^*$-embedded in $X$ if and only if any two completely separated sets in $S$ are completely separated in $X$.
The proof goes as usual, with the construction of a sequence of continuous functions that converges uniformly to the desired function. Then, after proving Urysohn's Lemma (for normal spaces), UT yields Tietze's Theorem (TT).
On the other hand, there are some ways to prove TT without mentioning uniform convergence at all. For instance, as done by Scott here.
Since UT is a generalization of TT, I wonder if there is a way to prove UT without using uniform convergence. I tried to adapt Scott's argument to the general setting of UT, but I could not avoid the need for normality.
 A: I wasn't able to get access to Scott's paper, so I instead adapted Mandelkern's proof of TT below.
One direction of UT is easy, so we only have to prove that if $A \subset X$ is such that any two completely separated sets in $A$ are completely separated in $X$, then $A$ is $C^*$-embedded in $X$ -- it suffices to show that any continuous $f : A \to [0, 1]$ extends to a continuous $F : X \to [0, 1]$. Mandelkern proves this instead starting with the assumption of normality, but notably normality is only invoked once: in the construction of a closed $H_n$ such that
$$A_{r_n} \cup \bigcup_{j \in J} H_j \subset H_n^\circ \subset H_n \subset U_{s_k} \cap \bigcap_{k \in K} H_k^\circ.$$ 
We can get around this by slightly altering the inductive argument. 
The proof uses an enumeration $\{(r_n, s_n)\}_{n=1}^\infty$ of the set $P = \{(r, s) \mid r, s \in \mathbb{Q}, 0 \leq r < s < 1\}$, and defines $A_r = f^{-1}([0, r])$ and $U_s = X \setminus f^{-1}([s, 1])$ for $r, s \in \mathbb{Q}$. The proof then inductively constructs a sequence $\{H_n\}$ of closed subsets of $X$ such that
$$A_{r_n} \subset H_n^\circ \subset H_n \subset U_{s_n} \quad \text{for each $n$}$$ and such that 
$$H_j \subset H_k^\circ \quad \text{whenever $j, k$ satisfy $r_j < r_k$ and $s_j < s_k$}.$$
We can alter the induction by strengthening the conditions on $H_n$: instead of the above, we require that $A_{r_n}$ is completely separated from $(H_n^\circ)^c$, $H_n$ is completely separated from $U_{s_n}^c$, and that for $j, k$ satisfying the above condition, $H_j$ is completely separated from $(H_k^\circ)^c$. Now in the revised induction, to construct $H_n$ we'll use the following lemma. 
Lemma. If $C_1, \dots, C_k$ and $D_1, \dots, D_\ell$ are such that each pair $C_i$ and $D_j$ are completely separated, then $C_1 \cup \cdots \cup C_k$ and $D_1 \cup \cdots \cup D_\ell$ are completely separated. 
Proof. Let $g_{ij} : X \to [0, 1]$ be such that $g_{ij}|C_i = 0$ and $g_{ij}|D_j = 1$. Defining $g_j = \prod_{i=1}^k g_{ij}$ for each $j$, it's clear that $g_j = 0$ on $C_1 \cup \cdots \cup C_k$ and $g_j = 1$ on $D_j$. Now defining $g = 1 - \prod_{j=1}^\ell (1 - g_j)$, we have that $g = 0$ on $C_1 \cup \cdots \cup C_k$ and $g = 1$ on $D_1 \cup \cdots \cup D_\ell$. 
Now, having constructed $H_k$ with the desired properties for $1 \leq k < n$, we proceed to construct $H_n$. As in the proof, define
$$J = \{j \mid j < n, r_j < r_n \text{ and } s_j < s_n\}$$
and 
$$K = \{k \mid k < n, r_n < r_k \text{ and } s_n < s_k\}.$$
By our induction, for each $j \in J$ and $k \in K$, $j$ and $k$ satisfy the condition, so $H_j$ and $(H_k^\circ)^c$ are completely separated in $X$. Also, for each $j \in J$, $H_j$ is completely separated from $U_{s_j}^c \supset U_{s_n}^c$, hence from $U_{s_n}^c$ as well. Similarly, for each $k \in K$, $(H_k^\circ)^c$ is completely separated from $A_{r_k} \supset A_{r_n}$, hence from $A_{r_n}$ as well. Finally, $A_{r_n} = f^{-1}([0, r_n])$ and $U_{s_n}^c = f^{-1}([s_n, 1])$ are completely separated in $A$ (by a truncated version of $f$), hence they are also completely separated in $X$ by assumption. 
Thus we can apply our lemma to the set $A_{r_n}$ together with all $H_j$ for $j \in J$ against the set $U_{s_n}^c$ together with all $(H_k^\circ)^c$ for $k \in K$. It follows that 
$$A_{r_n} \cup \bigcup_{j \in J} H_j \quad \text{and} \quad U_{s_n}^c \cup \bigcup_{k \in K} (H_k^\circ)^c$$
are completely separated in $X$ by some function $g : X \to [0, 1]$ (with $g = 0$ on the first set and $g = 1$ on the second). We can finish by picking $H_n = g^{-1}([0, 1/2])$; it is not hard to verify that all of the desired properties of $H_n$ hold from here. The rest of the proof is unchanged.
