An inner product on a space $V$ can be extended to its completion I am self studying from Reed and Simon's functional analysis book and I am working on the following problem.

(a) Let $V$ be an inner product space. Prove that the inner product can be extended to $\tilde{V}$ as follows: First show that if $x,y \in \tilde{V}$, $x_n, y_n \in V$ and $x_n \rightarrow x, y_n \rightarrow y$, then $(x_n, y_n)$ converges. Define $(x,y) = \lim_{n\rightarrow} (x_n, y_n)$ and show that it is independent of which convergent subsequences are chosen. Finally show that $(\cdot, \cdot)$ has the right properties.
(b) Prove the statement in (a) by applying the B.L.T. theorem twice.

My proof:
(a) Consider $n, m \geq N$
$$|(x_n - x_m, y_n - y_m)| \leq |(x_n -x + x - x_m, y_n-y+y-y_m)| \\
\leq |(x_n -x, y_n -y)| + |(x_n -x, y-y_m)| + |(x-x_m, y_n -y)| + |(x-x_m, y-y_m) \\ < \epsilon$$
I've omitted  some small details but the second to last inequality follows from the assumption of convergence. Thus $(x_n, y_n)$ is Cauchy, and since $\mathbb{R}$ is complete, it must converge to some limit $L \equiv (x,y)$.
To show that the limit is independent of the chosen convergent sequence, suppose that $\tilde{x}_n \rightarrow x$ and $\tilde{y}_n \rightarrow y$. Since each sequence was assumed to converge we have that:
$$\|x_{n} - x\| < \epsilon_1, \quad \forall n > N_1\\
\|\tilde{x}_{n} - x\| < \epsilon_2, \quad \forall n > N_2\\
\|y_{n} - y\| < \epsilon_3, \quad \forall n > N_3\\
\|\tilde{y}_{n} - y\| < \epsilon_4, \quad \forall n > N_4$$
Let $\frac{\sqrt{\epsilon}}{2} = \min\{\epsilon_1, \epsilon_2, \epsilon_3, \epsilon_4\}$ and $N = \max\{N_1, N_2, N_3, N_4\}$. Then for all $n \geq N$:
$$\big|(x_n - \tilde{x}_{n}, y_n-\tilde{y}_{n})\big| = \big|(x_n - x + x - \tilde{x}_{n}, y_n - y + y - \tilde{y}_{n})\big| \\ 
\leq \big|(x_n-x, y_n -y)| + \big|(x_n-x, y -\tilde{y}_{n})| + \big|(x-\tilde{x}_{n}, y_n -y)| + \big|(x-\tilde{x}_{n}, y -\tilde{y}_{n})|\\
\leq \|x_n -x\|\|y_n-y\| + \|x_n-x\|\|y-\tilde{y}_{n}\| + \|x-\tilde{x}_{n}\|\|y_n-y\| + \|x-\tilde{x}_{n}\|\|y-\tilde{y}_{n}\| \\
< \epsilon$$
and thus uniqueness has been shown. I've used the Cauchy-Schwartz inequality in the second to last inequality.
To show that the inner product has the right properties suppose $x_n \rightarrow x, y_n \rightarrow y, z_n \rightarrow z$:
Then $(x_n, x_n) \geq 0$ for all since $(\cdot, \cdot)$ is assumed to be an inner product, and since we showed that the inner product converges. $(x, x) \geq 0$ as well. The property that $(x,x) = 0$ if and only if $x = 0$ is obvious by the same reasoning.
We also have that $(x, \alpha(y + z)) = \lim_{n\rightarrow \infty}(x_n, \alpha(y_n + z_n)) = \lim_{n\rightarrow \infty} \big(\alpha(x_n, y_n) + \alpha(x_n, z_n)\big) = \alpha(x,y) + \alpha(x,z)$. Showing that $(x,y) = \overline{(y,x)}$ uses the same reasoning.
Hence $(\cdot, \cdot)$ is an inner product on the completion $\tilde{V}$.
(b) Fix $y \in V$. The operator $T_y$ defined by the inner product:
$$T_y: V \rightarrow \mathbb{R}\\
x \mapsto (x,y)$$
is clearly linear and bounded (by Cauchy-Schwartz). Since $V$ is a normed linear space and $\mathbb{R}$ is a complete normed linear space, we may extend $T_y$ to $\tilde{V}$ by the BLT theorem.
I am stuck here. My initial thought was to use the BLT theorem again but this time fixing $x$ to construct an operator $T_x$, and somehow composing the two operators, but $T_x$ will not be linear but rather conjugate linear and so I cannot use the BLT theorem.

Can anyone verify my proof? Apart from part (b), I am uncertain about (1) where one limit turns into two, but I do not know of a workaround to this.
 A: The argument in (a) is not good as stated, because you use the inner product in elements that are not in $V$; and the inequality you get is not what you need. You want to show that if $\{x_n\}$ and $\{y_n\}$ are Cauchy in $V$, then $\{(x_n,y_n)\}$ is Cauchy in $\mathbb C$. For this, you have to use Cauchy-Schwarz and the fact that Cauchy sequences in a metric space are bounded. So there exists $c\in\mathbb R$ such that $\|x_n\|\leq c$ and $\|y_n\|\leq c$ for all $n$. Then
\begin{align}
|(x_n,y_n)-(x_m,y_m)|&\leq |(x_n-x_m,y_n)|+|(x_m,y_n-y_m)|\\[0.3cm]
&\leq\|x_n-x_m\|\,\|y_n\|+\|x_m\|\,\|y_n-y_m\|\\[0.3cm]
&\leq c\,\|x_n-x_m\|+c\,\|y_n-y_m\|.
\end{align}
Since both sequences are Cauchy, there exists $N$ such that for all $n,m\geq N$ we have $\|x_n-x_m\|<\varepsilon/2c$ and $\|y_n-y_m\|<\varepsilon/2c$. Then, for all $n,m\geq N$,
$$
|(x_n,y_n)-(x_m,y_m)|<\varepsilon,
$$
which shows that $\{(x_n,x_m)\}$ is Cauchy. This allows us to define
$$
(x,y)=\lim_n(x_n,y_m). 
$$
Now if $x_n'\to x$ and $y_n'\to y$. The same inequalities as above give us
\begin{align}
|(x_n,y_n)-(x_n',y_n')|&\leq c\,\|x_n-x_n'\|+c\,\|y_n-y_n'\|\\[0.3cm]
&\leq c\|x_n-x\|+c\,\|x-x_n'\|+c\,\|y_n-y\|+c\,\|y-y_n'\|.
\end{align}
So $(x_n',y_n')$ converges to the same limit as $(x_n,y_n)$, showing that $(x,y)$ does not depend on the sequences used.
Your version of the uniqueness has the same problem as your existence, which is that you are using the inner product on elements of the completion before you know that said product exists.
Later, you say

The property that $(x,x)=0$ if and only if $x=0$  is obvious by the same reasoning.

I have no idea what you mean by this, so here is my take. If $x=0$, since $0\in V$, there is nothing to prove. Now, if $(x,x)=0$ for some $x$ in the completion, an argument is needed. Namely, we need to show that if $\{x_n\}\subset V$ and $(x_n,x_n)\to0$, then $x_n\to0$. This is when I would say that it is obvious: the equality $(x,x)=0$ gives us $\|x_n\|^2=(x_n,x_n)\to0$, which means that $x_n\to0$. Then $x=0$.
Once the above is established, you argument for the inner product properties is standard and correct, since the limit preserves those operations.
For part (b), it is weird that you use $\mathbb R$ but then are worried about the complex conjugate? I'll assume you meant $\mathbb C$. In that case, the argument with the BLT Theorem is almost as you say. First you get  $T_y:\tilde V\to \mathbb C$ with
$$
|T_y(x)|\leq \|y\|\,\|x\|,
$$
and $T_y(x)=(x,y)$ for all $x\in V$.
fix $x\in\tilde V$ and consider the linear operator $S_x: V\to\mathbb C$ given by
$$
S_x(y)=\overline{T_y(x)}.
$$
This operator is linear and bounded on $V$. Indeed, when $x\in V$ we have
\begin{align}
S_x(y_1+\alpha y_2)&=\overline{T_{y_1+\alpha y_2}(x)}=\overline{(x,y_1+\alpha y_2)}=(y_1+\alpha y_2,x)=(y_1,x)+\alpha(y_2,x)\\[0.3cm]
&=\overline{(x,y_1)}+\alpha\overline{(x,y_2)}=S_x(y_1)+\alpha S_x(y_2). 
\end{align}
So $S_x$ is linear when $x\in V$. When $x\in\tilde V$, we have $x=\lim_nx_n$ for some sequence $\{x_n\}\subset V$. Then
$$
S_x(y)=\overline{T_y(x)}=\lim_n\overline{T_y(x_n)}=\lim_nS_{x_n}(y), 
$$
and the linearity of $S_x$ follows from the linearity of the limit and that of $S_{x_n}$. And for boundedness,
$$
|S_x(y)|=|T_y(x)|\leq\|x\|\,\|y\|,
$$
so $S_x$ is bounded. By BLT Theorem, it extends to all of $\tilde V$. Once $S_x(y)$ is defined for all $x,y\in\tilde V$, we can define
$$
(x,y)=\overline{S_x(y)},\qquad\qquad x,y\in\tilde V. 
$$
This function satisfies $|(x,y)|\leq\|x\|\,\|y\|$ and it agrees with the original inner product when $x,y\in V$.
