For metric spaces you can do another approach as well (yours is very specific for the reals), following Engelking exercise 4.1F, but the result is due to Hausdorff.
Let $(X,d)$ be metric and suppose $A$ is closed in $X$, and $f: A \rightarrow [0,1]$ is continuous. Then define
$$F(x) = \begin{cases} f(x) & \text{if } x \in A \\ \inf{\left\{f(a) + \frac{d(x,a)}{d(x,A)} - 1\,:\,a \in A\right\}} & \text{if } x \in X \smallsetminus A\end{cases}$$
and show that $F$ is continuous.
(For unbounded functions use the $\arctan(f)$ to make it bounded first.)
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.
Best Answer
There is a proof that uses a more sophisticated version of the ideas used in the usual proof of the Uryson lemma. You can find it in my old paper ‘A “More Topological” Proof of the Tietze-Urysohn Theorem’, The American Mathematical Monthly, Vol. $85$, No. $3$ (Mar., $1978$), pp. $192$-$193$,.