Verify: Prove the Continuous Extension Theorem Let $A$ be a bounded subset of $\mathbb{R}$ and $f: A \to \mathbb{R}$ be continuous.  Prove that $f$ is uniformly continuous on $A$ if and only if it is possible to construct $\hat f: \bar A \to \mathbb{R}$ as a continuous extension of $f$ to $\bar A$ (the closure of $A$).
Note: Proofs of the Continuous Extension Theorem are readily available.  This question asks to critique my proof as well as several questions below.
Lemma: Let $f: A \subseteq \mathbb{R} \to \mathbb{R}$ be uniformly continuous, and let $(x_n) \in A$ converge to $x$ (not necessarily in $A$).  Then $(f(x_n))$ converges.  Moreover, if $(x'_n) \in A$ converges to $x$, then $(f(x'_n))$ converges to the limit of $(f(x_n))$.
Proof: Since $f$ is uniformly continuous, for every $\varepsilon > 0$, there exists $\delta > 0$ such that $|x_n - x_m| < \delta \implies |f(x_n) - f(x_m)| < \varepsilon$.  Additionally, since $(x_n)$ converges, it is a Cauchy sequene, and there exists $N$ such that for all $n, m \geq N, |x_n - x_m| < \delta$, so that $|f(x_n) - f(x_m)| < \varepsilon$.  Thus, $(f(x_n))$ is a Cauchy sequence and converges.
Suppose $(x'_n) \in A$ also converges to $x$.  Then $|x_n - x'_n| \to 0$, and since $f$ is uniformly continuous, $|f(x_n) - f(x'_n)| \to 0$, so $\lim f(x'_n) = \lim f(x_n)$.
Main Proof: Clearly $\bar A$ is compact, and so if continuous $\hat f$ exists, then $\hat f$ as well as its restriction $f$ are uniformly continuous.  We now prove the converse (that if $f$ is uniformly continuous, we can extend it to continuous $\hat f: \bar A \to \mathbb{R}$) by constructing $\hat f$ as follows.
As an extension of $f$, $\hat f(x) := f(x)$ for all $x \in A$.  Recall that for all $x \in \bar A$, either $x \in A$ or $x$ is a limit point of $A$.  Therefore, for any $x$ in $\bar A$ but not in $A$, we can construct a sequence $(x_n) \in A$ converging to $x$.  By the lemma, $f(x_n)$ will converge to same limit for all such sequences, and so we define $\hat f(x) = \lim f(x_n)$.
Observe that $\hat f$ is continuous, since for any $(x_n) \in \bar A$ converging to $x$, $\lim \hat f(x_n) = \lim f(x_n) = \hat f(x)$ by construction.

Discussion

*

*Is the proof correct and rigorous?

*Is it well written? How can the exposition be improved?

*This proof is for any bounded set $A$ and its closure.  I've seen the theorem presented only for open intervals, but I believe, as written, it applies to all bounded sets.  Am I right?

*The lemma asserts not only that $(f(x_n))$ converges, but that it does so to the same limit for all $(x_n) \to x$.  I believe this part is necessary and sufficient to establish continuity of $\hat f$.  Am I correct?

*Using #4, I encapsulated most of the complexity into the lemma, with the main proof following almost immediately.  Does this structure clarify the exposition?

*The lemma holds only for uniformly continuous functions, since $(x_n)$ may converge to a limit outside of $A$.  However, if $A$ is closed, a similar property holds for any continuous function:  Since $(x_n) \in A \to x \in A$, then $(f(x_n)) \to f(x)$.  Thus, continuous functions on closed domains are Cauchy-continuous.  Is this of significance?

 A: Here is a proposal of improvement, to "consolidate" my comments. (Against my taste, I stay with numerical functions, to respect your choice.)
Let $A$ be a subset of $\Bbb R.$
Definition A map $f:A\to\Bbb R$ is:

*

*uniformly continuous if [...]


*Cauchy-continous if [...]
Three obvious remarks
R1) Uniform continuity and Cauchy-continuity are preserved by restriction.
R2) Any uniformly continuous function is Cauchy-continuous.
R3) Any Cauchy-continuous function is continuous.
The converse of R2 and R3 are false [examples
...] but:
[Heine-Cantor theorem...]
Two more obvious remarks
R4) Any continuous function on a closed subset of $\Bbb R$ is Cauchy-continuous (this may be used to certify that the extension $\hat f$ of the theorem below is not only continuous but Cauchy-continuous).
R5) (not so interesting, from my point of view) As a consequence of Heine-Cantor theorem and of R1, if $A$ is bounded and $f$ extends to a continous map $\hat f:\bar A\to\Bbb R,$ then $f$ is uniformly continuous.
We shall now state a "strong converse" of R5, which does not require $A$ to be bounded.
Continuous extension theorem Let $f:A\to\Bbb R$ be Cauchy-continuous. Then, $f$ admits a continuous extension $\hat f:\bar A\to\Bbb R.$
Proof
Let $x\in\bar A.$ For any sequence $(a_n)$ of elements of $A$ converging to $x,$ $(f(a_n))$ is a Cauchy sequence by R2, hence admits a limit $y.$ Moreover, $y$ does not depend on the choice of $(a_n)$ because if $(a'_n)$ is another such sequence, then $(a_1,a'_1,a_2,a'_2,\dots)$ also is, hence (by what we just showed) its image by $f$ admits a limit, which proves that $(f(a'_n))$ has the same limit $y$ as $(f(a_n)).$ This allow us to define $\hat f(x)$ as being this common limit $y.$
Note that $\hat f$ is an extension of $f$ (if $x\in A$, we may apply the definition of $\hat f(x)$ with the constant sequence $a_n=x$).
Let us now prove that $\hat f$ is continuous. Let $(x_n)$ be a converging sequence in $\bar A,$ and $x$ be its limit. For each $n\in\Bbb N,$ by definition of $\hat f(x_n),$ there exists an element $a_n\in A$ such that $|x_n-a_n|$ and $|\hat f(x_n)-f(a_n)|$ are both $<\frac1n.$ Using such a sequence $(a_n)$, we deduce: $\hat f(x)=\lim f(a_n)=\lim\hat f(x_n).$
