Recall that any metric space $M$ has a metric $d$ defined on it where:
$$d:M\times M\to \mathbb R_{\geq 0}$$
and $d$ fulfills the following axioms:
Symmetry:
$$d(x,y) = d(y,x)$$
Non-negativity
$$d(x,y)\geq 0\qquad d(x,y)=0\iff x = y$$
Triangle inequality:
$$d(x,y)\leq d(x,z)+d(z,y)$$
We can apply the triangle inequality twice to $d(x_n,y_n)$ as follows:
$$d(x_n,y_n) \leq d(x_n,y)+d(y,y_n)$$
We also have that $$d(x_n,y)\leq d(x_n,x)+d(x,y)$$
We can combine these to get that:
$$d(x_n,y_n)\leq d(x,x_n)+d(x,y)+d(y,y_n)$$
Here I implicitly used the symmetry condition to say that $d(x,x_n)=d(x_n,x)$. It wasn't required, but it's good to recognize that these are equal.
We can interchange $x_n$ and $x$ and $y_n$ and $y$ because throughout this argument we've been treating them as points, and "forgetting" that they are sequences. We could have used points called $a,b,c,d$ and still obtained a valid relation between them. So, interchange the points is just saying "this argument is true for any $4$ points, and as we don't need to repeat it, we'll just take the end result and modify it to how we want it").
Suppose that $x = \sum\limits_{k = 1}^\infty \alpha_k e_k$. Then for all $\epsilon > 0$, there exists $n$ s.t. $|x - \sum\limits_{k = 1}^n \alpha_k e_k| < \epsilon$. And clearly, $\sum\limits_{k = 1}^n \alpha_k e_k \in M$. Then $x$ is a limit point of $M$; that is, $x \in \overline{M}$.
Claim: for all $x$, the quantities $\alpha_1, ..., \alpha_k$ given as above minimise $|x - \sum\limits_{k = 1}^n x_k e_k|$. Proof: fairly trivial. We will use claim twice in a moment.
Conversely, suppose $x \in \overline{M}$. We wish to show that $x = \sum\limits_{k = 1}^\infty \alpha_k e_k$ where $\alpha_k$ is defined as above. Given $\epsilon > 0$, take some element $m \in M$ s.t. $|x - m| < \epsilon$ (this is possible because $x$ is a limit point of $M$). Then for some $n$, we may write $m = \sum\limits_{k = 1}^n m_k e_k$ for some complex numbers $m_k$. But it can be shown that $\alpha_1, ..., \alpha_n$ minimise $|x - \sum\limits_{k = 1}^n m_k e_k|$ Then $|x - \sum\limits_{k = 1}^n \alpha_k e_k| \leq |x - \sum\limits_{k = 1}^n m_k e_k| < \epsilon$. Now suppose that we have some $n' \geq n$. Then we see that $|x - \sum\limits_{k = 1}^{n'} \alpha_k e^k| \leq |x - \sum\limits_{k = 1}^{n'} \beta_k e_k| = |x - \sum\limits_{k = 1}^n \alpha_k e_k| < \epsilon$, where we define $\beta_k = 0$, $k > n$ and $\beta_k = \alpha_k$, $k \leq n$. We again cite our above claim to prove this.
Then by the $n-\epsilon$ definition of limit, we have $x = \lim\limits_{n' \to \infty} \sum\limits_{k = 1}^{n'} \alpha_k e_k = \sum\limits_{k = 1}^\infty \alpha_k e_k$, as required.
Best Answer
I’ll expand on Kreyszig’s argument. I don’t have the book, so it’s possible that some of what I’m about to do duplicates earlier parts of his proof.
We start with two completions of $\langle X,d\rangle$, $\langle\hat X,\hat d\rangle$ and $\langle\tilde X,\tilde d\rangle$. This means that there are $W\subseteq\hat X$ and $\tilde W\subseteq\tilde X$ such that
We want to show that there is an isometry $h:\tilde X\to\hat X$ such that $h\upharpoonright\tilde W=T\circ\tilde T^{-1}:\tilde W\to W$.
Let $\tilde x\in\tilde X$. $\tilde W$ is dense in $\tilde X$, so for each $n\in\Bbb Z^+$ there is a point $\tilde x_n\in B_{\tilde d}\left(\tilde x,\frac1n\right)\cap\tilde W$. Clearly the sequence $\langle\tilde x_n:n\in\Bbb Z^+\rangle$ converges to $\tilde x$ in $\tilde X$. For each $n\in\Bbb Z^+$ let $x_n=\tilde T^{-1}(\tilde x_n)$ and $\hat x_n=T(x_n)\in\hat X$. $\tilde T$ and $T$ are isometries, so for all $m,n\in\Bbb Z^+$ we have
$$\hat d(\hat x_m,\hat x_n)=\tilde d(\tilde x_m,\tilde x_n)\;.\tag{1}$$
The sequence $\langle\tilde x_n:n\in\Bbb Z^+\rangle$ is convergent in $\tilde X$, so it’s Cauchy, and it follows immediately from $(1)$ that $\langle\hat x_n:n\in\Bbb Z^+\rangle$ is Cauchy in $\hat X$. And $\hat X$ is complete, so $\langle\hat x_n:n\in\Bbb Z^+\rangle$ must converge to some $\hat x\in\hat X$.
Suppose that $\langle\tilde x_n':n\in\Bbb Z^+\rangle$ is another sequence in $\tilde W$ converging to $\tilde x$. For each $\epsilon>0$ there is an $m_\epsilon\in\Bbb Z^+$ such that $\tilde d(\tilde x_n,\tilde x),\tilde d(\tilde x_n',\tilde x)<\frac{\epsilon}2$ whenever $n\ge m_\epsilon$, and hence
$$\tilde d(\tilde x_n,\tilde x_n')\le\tilde d(\tilde x_n,\tilde x)+\tilde d(\tilde x,\tilde x_n')<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$
whenever $n\ge m_\epsilon$. For each $n\in\Bbb Z^+$ let $x_n'=\tilde T^{-1}(\tilde x_n')\in X$ and $\hat x_n'=T(x_n')\in\hat X$. Again using the fact that $\tilde T$ and $T$ are isometries, we have $\hat d(\hat x_n,\hat x_n')=\tilde d(\tilde x_n,\tilde x_n')$ for each $n\in\Bbb Z^+$, and hence $\hat d(\hat x_n,\hat x_n')<\epsilon$ whenever $n\ge m_\epsilon$.
Let $\epsilon>0$. Since $\langle\hat x_n:n\in\Bbb Z^+\rangle$ converges to $\hat x$ in $\hat X$, there is a $k_\epsilon$ such that $\hat d(\hat x_n,\hat x)<\frac{\epsilon}2$ whenever $n\ge k_\epsilon$. Thus,
$$\hat d(\hat x_n',\hat x)\le\hat d(\hat x_n',\hat x_n)+\hat d(\hat x_n,\hat x)<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$
whenever $n\ge\max\{k_\epsilon,m_{\epsilon/2}\}$, and $\langle\hat x_n':n\in\Bbb Z^+\rangle$ also converges to $\hat x$.
This shows that the map $h:\tilde X\to\hat X$ defined by $h(\tilde x)=\hat x$ is well-defined: no matter which sequence in $\tilde W$ converging to $\tilde x$ I use to define $\hat x$, I get the same point of $hat X$. Moreover, if $\tilde x\in\tilde W$, then I can let $\tilde x_n=\tilde x$ for each $n\in\Bbb Z^+$ and see that $h(\tilde x)=T\left(\tilde T^{-1}(\tilde x)\right)$, so that $h\upharpoonright\tilde W=T\circ\tilde T^{-1}$, as desired.
Now we have to check that $h$ is an isometry.
Let $\tilde x$ and $\tilde y$ be any two points in $\tilde X$, and let $\langle\tilde x_n:n\in\Bbb Z^+\rangle$ and $\langle\tilde y_n:n\in\Bbb Z^+\rangle$ be sequences in $\tilde W$ converging to $\tilde x$ and $\tilde y$, respectively. Then the sequence $\big\langle\langle\tilde x_n,\tilde y_n\rangle:n\in\Bbb Z^+\big\rangle$ converges to $\langle\tilde x,\tilde y\rangle$ in the space $\tilde X\times\tilde X$. The metric $\tilde d$ is a continuous function from $\tilde X\times\tilde X$ to $\Bbb R$, so the sequence $\langle\tilde d(\tilde x_n,\tilde y_n):n\in\Bbb Z^+\rangle$ converges to $\tilde d(\tilde x,\tilde y)$ in $\Bbb R$.
As before, for each $n\in\Bbb Z^+$ let $x_n=\tilde T^{-1}(\tilde x_n)\in X$, and let $\hat x_n=T(x_n)\in W$, so that $\langle\hat x_n:n\in\Bbb Z^+\rangle$ converges to $h(\tilde x)$ in $\hat X$. Similarly, for each $n\in\Bbb Z^+$ let $y_n=\tilde T^{-1}(\tilde y_n)\in X$ and $\hat y_n=T(y_n)\in W$, so that $\langle\hat y_n:n\in\Bbb Z^+\rangle$ converges to $h(\tilde y)$.
The same argument that I used to show that $\langle\tilde d(\tilde x_n,\tilde y_n):n\in\Bbb Z^+\rangle$ converges to $\tilde d(\tilde x,\tilde y)$ in $\Bbb R$ can be used to show that $\langle\hat d(\hat x_n,\hat y_n):n\in\Bbb Z^+\rangle$ converges to $\hat d(\hat x,\hat y)$ in $\Bbb R$. But $\tilde d(\tilde x_n,\tilde y_n)=d(x_n,y_n)=\hat d(\hat x_n,\hat y_n)$ for each $n\in\Bbb Z^+$, so
$$\hat d\big(h(x),h(y)\big)=\lim_{n\to\infty}\hat d(\hat x_n,\hat y_n)=\lim_{n\to\infty}\tilde d(\tilde x_n,\tilde y_n)=\tilde d(\tilde x,\tilde y)\;,$$
and $h$ is indeed an isometry.
The last step is to show that $h$ maps $\tilde X$ onto $\hat X$. Let $\hat x\in\hat X$ be arbitrary. $W$ is dense in $\hat X$, so there is a sequence $\langle\hat x_n:n\in\Bbb Z^+\rangle$ in $W$ converging to $\hat x$. Use the ideas above to show that the sequence
$$\left\langle\tilde T\left(T^{-1}(\hat x_n)\right):n\in\Bbb Z^+\right\rangle$$
in $\tilde W$ is Cauchy and therefore converges to some $\tilde x\in\tilde X$; then show that $\hat x=h(\tilde x)$.