Prob. 11, Chap. 4 in Baby Rudin: uniformly continuous extension from a dense subset to the entire space

Here is Prob. 11, Chap. 4 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f$ is a uniformly continuous mapping of a metric space $X$ into a metric space $Y$ and prove that $\left\{ f\left( x_n \right) \right\}$ is a Cauchy sequence in $Y$ for every Cauchy sequence $\left\{ x_n \right\}$ in $X$. Use this result to give an alternative proof of the theorem stated in Exercise 13.

Now here is Prob. 13, Chap. 4 in Baby Rudin, 3rd edition:

Let $E$ be a dense subset of a metric space $X$, and let $f$ be a unifromly continuous real function defined on $E$. Prove that $f$ has a continuous extension from $E$ to $X$. ... (Uniqueness follows from Exercise 4.) ...

And, here is Prob. 4, Chap. 4:

Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $E$ be a dense subset of $X$. Prove that $f(E)$ is dense in $f(X)$. If $g(p) = f(p)$ for all $p \in E$, prove that $g(p) = f(p)$ for all $p \in X$. (In other words, a continuous mapping is determined by its values on a dense subset of its domain.)

My effort:

Since $f$ is uniformly continuous on $X$, corresponding to every real number $\varepsilon > 0$, we can find a real number $\delta > 0$ such that $$d_Y\left(f(x), f(y)\right) < \varepsilon$$ for all points $x, y \in X$ which satisfy $$d_X(x,y)<\delta.$$

Now since $\left\{ x_n \right\}$ is a Cauchy sequence in $\left( X, d_X \right)$, therefore corresponding to the real number $\delta$ in the preceding paragraph we can find a natural number $N$ such that $$d_X\left(x_m, x_n \right)< \delta \ \mbox{ for any natural numbers } m \mbox{ and } n \mbox{ such that } m > N \mbox{ and } n > N.$$

So from the preceding two paragraphs we can conclude that $$d_Y\left( f\left(x_m\right), f\left(x_n\right) \right) < \varepsilon \ \mbox{ for any natural numbers } m \mbox{ and } n \mbox{ such that } m > N \mbox{ and } n > N,$$ from which it follows that $\left\{ f\left(x_n\right) \right\}$ is a Cauchy sequence in $\left( Y, d_Y \right)$.

Now let $E$ be a dense set in $\left(X, d_X \right)$, $\left( Y, d_Y \right)$ be a complete metric space, and $f$ be a uniformly continuous mapping of $E$ into $Y$.

We now show that there exists a unique uniformly continuous mapping $g$ of $X$ into $Y$ such that $g(p) = f(p)$ for all $p \in E$.

Let $p \in X$. Since $E$ is dense in $X$, we can find a sequence $\left\{p_n\right\}$ in $E$ converging to the point $p$ in $\left( X, d_X \right)$.

Now as the sequence $\left\{ p_n \right\}$ is a Cauchy sequence in $E$ and as $f$ is uniformly continuous, so the image sequence $\left\{ f\left(p_n \right)\right\}$ is a Cauchy sequence in the complete metric space $Y$, and so this sequence converges to some point $q$ in $Y$.

We now show that this point $q$ is independent of the particular sequence $\left\{ p_n \right\}$ in $E$ which converges to $p$. For this, let $\left\{ p^\prime_n \right\}$ be another sequence in $E$ converging in $X$ to the point $p$. Then as before the image sequence $\left\{ f\left(p^\prime_n\right)\right\}$ converges in $Y$ to some point $q^\prime$. Now consider the sequence $\left\{ x_n \right\}$ in $E$ defined as follows: Let $$x_n = \begin{cases} p_{\frac{n}{2}} \ \mbox{ if $n$ is even}; \\ p^\prime_{\frac{n+1}{2}} \ \mbox{ if $n$ is odd}. \end{cases}$$

This sequence too converges to point $p$ in $X$ and is therefore Cauchy and so its image sequence $\left\{ f\left( x_n \right) \right\}$ is also a Cauchy sequence in $Y$, which is a complete metric space; so the sequence $\left\{ f\left( x_n \right) \right\}$ converges in $Y$ to some point $y$. Therefore every subsequence of $\left\{ f\left( x_n \right) \right\}$ also converges to $y$.

But we note that, $$p_n = x_{2n} \ \mbox{ and } p^\prime_n = x_{2n-1} \ \mbox{ for all } n \in \mathbb{N}.$$ So $$q = \lim_{n \to \infty} f\left(p_n \right) = \lim_{n \to \infty} f\left( x_{2n}\right) = y,$$ where the limit is in $Y$, and similarly $$q^\prime = \lim_{n\to\infty} f\left(p_{2n-1}\right) = y.$$ Thus $q^\prime = q$.

Now let's define the mapping $g$ of $X$ into $Y$ as follows: For any point $p \in X$, let $$\tag{1} g(p) = \lim_{n \to \infty} f \left(p_n \right),$$ the limit being in the metric space $\left(Y, d_Y \right)$, where $\left\{ p_n \right\}$ is any sequence in $E$ such that $$\tag{2} \lim_{n \to \infty} p_n = p$$ in $\left( X, d_X \right)$.

If $p \in E$, then we can take $p_n = p$ for all $n$ so that $$g(p) = \lim_{n \to \infty} f\left(p_n\right) = f(p).$$ Thus $g$ is an extension of $f$ from $E$ to $X$.

Let $\varepsilon > 0$ be a given real number. Since $f$ is uniformly continuous on $E$, we can find a real number $\delta > 0$ such that $$d_Y\left( f(p), f(q) \right) < \frac{\varepsilon}{3}$$ for all points $p$ and $q$ in $E$ for which $$d_X\left(p, q\right) < \delta. $$

Now let $u$ and $v$ be any two points of $X$ which satisfy $$d_X\left(u, v\right) < \frac{\delta}{3}.$$ Then since $E$ is dense in $X$, we can find points $u^\prime$ and $v^\prime$ in $E$ such that $$\tag{3} d_X \left( u, u^\prime \right) < \frac{\delta}{3} \mbox{ and } d_Y\left(g(u), f(u^\prime) \right) < \frac{\varepsilon}{3},$$ and also $$\tag{4} d_X \left( v, v^\prime \right) < \frac{\delta}{3} \mbox{ and } d_Y\left( g(v), f(v^\prime) \right) < \frac{\varepsilon}{3}.$$ This is possible in view of (1) and (2) above.

Then we note that $$d_X \left( u^\prime, v^\prime \right) \leq d_X\left(u^\prime, u \right) + d_X\left(u, v\right) + d_X\left(v, v^\prime\right) < \frac{\delta}{3} + \frac{\delta}{3} + \frac{\delta}{3} = \delta,$$ which implies that $$ \tag{5} d_Y\left( f\left(u^\prime \right), f\left(v^\prime\right) \right) < \frac{\varepsilon}{3},$$ and therefore from (3), (4), and (5) we can conclude that $$ \begin{align} d_Y\left( g\left(u\right), g\left(v\right) \right) &\leq d_Y\left( g\left(u \right), f\left(u^\prime \right) \right) + d_Y\left( f\left(u^\prime\right), f\left(v^\prime\right) \right) + d_Y\left( f\left(v^\prime\right), g\left(v\right) \right) \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} \\ &= \varepsilon. \end{align} $$

Let us take a real number $\eta$ such that $$0 < \eta < \frac{\delta}{3}.$$

Thus, corresponding to every real number $\varepsilon > 0$, we can find a positive real number $\eta$ such that $$d_Y \left( g(u), g(v) \right) < \varepsilon$$ for all points $u$ and $v$ in $X$ for which $$ d_X \left(u, v \right) < \eta.$$ Hence $g$ is uniformly continuous on $X$.

Let $g^\prime$ be (also) a (uniformly) continuous extension of $f$ from $E$ to $X$.

Since $g^\prime(p) = f(p) = g(p)$ for all $p \in E$ and since $E$ is dense in $X$, therefore by virtue of the conclusion in Prob. 4 we can conclude that $g^\prime(p) = g(p)$ for all $p \in X$.

Now I know that the first part of the above solution, where we are required to show that $\left\{ f\left(x_n\right) \right\}$ is a Cauchy sequence in $Y$ for every Cauchy sequence $\left\{ x_n \right\}$ in $E$, is correct, isn't it?

What about the second part? Is the result I've stated correct? If so, then is my proof correct also?

Your proof that such a function preserves Cauchy sequences is correct. Your statement and proof that uniformly continuous functions defined on dense subsets have uniformly continuous extensions is also correct.

Please add more context such as where exactly your doubts lie, and what they are. I don't feel like questions of this nature are necessarily appropriate when the answer can be as short as "yes" or "no."