In the Wikipedia's page of Tietze theorem is written that the first proof of this theorem was found for finite dimensional real vector spaces by Brouwer and Lebesgue, than for metric spaces by Tietze and finally for normal spaces by Urysohn.
My question is if there exist proofs of this theorem in the metric spaces setting that are easier than the general one (as for the Urysohn lemma) or even in the case of finite dimensional real vector space.
There is a "simpler" proof of the Tietze extension theorem for metric spaces, due to Hausdorff (1919): if $(X,d)$ is a metric space, $A\neq\varnothing$ is closed and $f:A\rightarrow[0,1]$ is continuous, then $\tilde{f}:X\rightarrow[0,1]$ given by $$\tilde{f}(p)=\begin{cases} f(p) & (p\in A) \\ \inf\Big\{f(q)+\frac{d(p,q)}{d(p,A)}-1\ \Big|\ q\in A\Big\} & (p\not\in A) \end{cases}\ ,\quad d(p,A)=\inf\{d(p,q)\mid q\in A\}$$ is continuous and clearly extends $f$. The quotes are due to the fact that the formula for the extension is more or less explicit (in fact, it reminds one of McShane's extension theorem for Lipschitz functions), but whether this proof can be considered simpler than Tietze's or not is up to discussion since it involves a fair amount of "epsilon-delta management".
It is easy to verify that $\tilde{f}$ takes values in $[0,1]$. Indeed, we have that $\frac{d(p,q)}{d(p,A)}-1\geq 0$ and $\tilde{f}(p)\leq\frac{d(p,q)}{d(p,A)}$ for all $p\in X\smallsetminus A$, $q\in A$, hence $\tilde{f}(p)\in[0,1]$ for all such $p$. The core of Hausdorff's proof is to establish the continuity of $\tilde{f}$. It follows from the definition of infimum of a nonvoid subset of $\mathbb{R}$ that $$\tilde{f}(p)\leq f(q)+\frac{d(p,q)}{d(p,A)}-1\ ,\quad p\in X\smallsetminus A\ ,\,q\in A$$ and $$\forall\delta>0\ ,\,p\in X\smallsetminus A\ \exists\ \!q\in A\text{ such that }f(q)+\frac{d(p,q)}{d(p,A)}-1<\tilde{f}(p)+\delta\ .$$ Particularly, the last assertion above entails that for such $q$ we have, since $\tilde{f}\leq 1$, that $$d(p,q)<(2+\delta)d(p,A)\ .$$ First we will establish continuity of $\tilde{f}$ in $A$. If $0<\delta<1$ and $p\in X\smallsetminus A$ is such that $d(p,A)<\frac{\delta}{4}$, we can find:
so that $$\tilde{f}(p)\leq f(q')+\frac{d(p,q')}{d(p,A)}-1<f(q')+\delta$$ and therefore $$f(q)-\delta<\tilde{f}(p)<f(q')+\delta\ .$$ Now, given $q_0\in A$, $\epsilon>0$, set $0<\delta<\min(\frac{\epsilon}{2},1)$ so that for all $r\in A$ with $d(r,q_0)<\delta$ we have $|f(r)-f(q_0)|<\frac{\epsilon}{2}<\epsilon-\delta$. If $d(p,q_0)<\frac{\delta}{4}$, we have that $q,q'$ as above satisfy $d(q,q_0)<\delta$ and $d(q',q_0)<\frac{3\delta}{4}$ by the triangle inequality, hence $|f(q)-f(q_0)|,|f(q')-f(q_0)|<\epsilon-\delta$ and therefore $$\begin{split}-\epsilon &<(\tilde{f}(p)-f(q))+(f(q)-f(q_0)) \\&=\tilde{f}(p)-f(q_0)\\&=(\tilde{f}(p)-f(q'))+(f(q')-f(q_0))\\ &<\epsilon\ .\end{split}$$ This establishes the continuity of $\tilde{f}$ in $q_0$. Since $q_0\in A$ was arbitrary, we conclude that $\tilde{f}$ is continuous in $A$. It remains to prove the continuity of $\tilde{f}$ in $X\smallsetminus A$ - let then $p_0,p\in X\smallsetminus A$, $0<\delta<\min(1,\frac{d(p_0,A)}{2})$ such that $d(p,p_0)<\delta$ and $q\in A$ such that $f(q)+\frac{d(p,q)}{d(p,A)}-1<\tilde{f}(p)+\delta$, hence $d(p,q)<3d(p,A)$. This entails that $$\begin{split} \tilde{f}(p_0) & \leq f(q)+\frac{d(p_0,q)}{d(p_0,A)}-1 \\ & <\tilde{f}(p)+\delta+\frac{d(p_0,q)}{d(p_0,A)}-\frac{d(p,q)}{d(p,A)} \\ &=\tilde{f}(p)+\delta+\frac{d(p_0,q)-d(p,q)}{d(p_0,A)}+d(p,q)\left(\frac{1}{d(p_0,A)}-\frac{1}{d(p,A)}\right) \\ &< \tilde{f}(p)+\delta\left(1+\frac{4}{d(p_0,A)}\right)\ .\end{split}$$ Notice that the inequality $\delta<\frac{d(p_0,A)}{2}$ was not used above, and the remaining hypotheses are symmetric in $p_0,p$. Hence, exchanging the roles of $p_0$ and $p$ (with a different $q$, adapted to $p_0$ instead of $p$), except on the choice of $\delta$, and then exploiting the fact that $$d(p,A)\geq d(p_0,A)-d(p,p_0)>d(p_0,A)-\delta>\frac{d(p_0,A)}{2}$$ by the triangle inequality, we obtain as well that $$ \tilde{f}(p)<\tilde{f}(p_0)+\delta\left(1+\frac{4}{d(p,A)}\right)<\tilde{f}(p_0)+\delta\left(1+\frac{8}{d(p_0,A)}\right)\ .$$ Both inequalities together establish the continuity of $\tilde{f}$ at $p_0$, as desired. Once more, $p_0\in X\smallsetminus A$ is arbitrary, hence $\tilde{f}$ is continuous in the whole of $X$.
The above proof can easily be modified to allow $f$ (and $\tilde{f}$) to assume values in any given compact interval of $\mathbb{R}$.
We actually did this in some form in an undergrad lecture on measure theory. To be precise we showed the following result (which was easier then the normal result^^):
Let $K$ be a compact metrical space, $C\subseteq K$ closed, and $f\colon C \to [-1,1]$. Then there exists $F\colon K \to [-1,1]$ such that $F|_C = f$.
We did this in multiple steps ($K$ being always compact, $A, B$ and $C$ being disjoint closed subsets):
1) Show that for all $A$, $B$ there is a continuous function $f\colon K \to [-1,1]$ and $g|_A = -1$, $g|_B = 1$. This can be done over the distance of a point from $A$ and $B$, using that both are closed and $K$ is compact for finiteness.
2) Let $f\colon C\to [-1,1]$ be continuous. Then there is a continuous $g\colon K\to [-2/3, 2/3]$ such that $|f-g| < 2/3 $ on $C$. For this set $A=f^{-1}([-1,-2/3])$ and $B=f^{-1}([2/3, 1])$ and apply 1) (and probably scale)
3) Show the result by building a sequence of functions $g_n$ ($g_0=f$) by repeatedly applying 2), always scaling with $2/3$ and taking the sum $\sum_{n=1}^\infty (2/3)^n g_n$.
