Your proof of (1) is incorrect. You start with a Cauchy sequence $\{T_n\} \subset B(V,V')$, and prove that it has a limit in $V$?!
What you want to do (and I suspect this is what you are trying to do) is as follows :
(1) For each $x\in V$, you check that $\{T_n(x)\} \subset V'$ is a Cauchy sequence. Since $V'$ is a Banach space, this sequence as a limit, which you can denote by $\alpha_x$.
(2) Now check that the elements $\{\alpha_x : x\in V\}$ satisfy the properties of a linear transformation. ie for any $x,y \in V$ and $c\in \mathbb{K}$,
$$
\alpha_{x+y} = \alpha_x + \alpha_y
$$
$$
\alpha_{cx} = c\alpha_x
$$
(3) So define $T : V \to V'$ by $T(x) = \alpha_x$, and check that $T$ defines a bounded linear map.
Now conclude that $T_n \to T$ in the operator norm defined on $B(V,V')$.
Expanding on your thoughts above, assume that $Y$ is a Banach space, then we would like to show that $B(X,Y)$ is a Banach space. Given a Cauchy sequence $\{ B_k \}_{k=1}^{\infty}$ in $B(X,Y)$, then for each $x \in X$, $\{ B_k x \}_{k=1}^{\infty}$ is indeed a Cauchy sequence in $Y$. Thus, there exists a limit $y \in Y$ such that $\| B_k x - y \|_Y \to 0$. However, it is not clear that $y =Bx$ for some $B \in B(X,Y)$.
To check this, define a function $B \colon X \to Y$ by $Bx = y$, where $y$ is obtained from $x$ in the manner above. We must check that this function $B$ is a bounded linear operator. For linearity, take $\alpha,\beta \in \mathbb{R}$ and $x_1,x_2 \in X$, then
\begin{align}
& \| B(\alpha x_1 + \beta x_2) - \alpha B(x_1) - \beta B(x_2) \|\\
&\leq \| B(\alpha x_1 + \beta x_2) - B_k(\alpha x_1 + \beta x_2) \|
+ \|\alpha B_k(x_1) - \alpha B(x_1) \| + \| \beta B_k(x_2) - \beta B(x_2) \|,
\end{align}
and all of these terms go to zero as $k \to \infty$. Therefore, $B(\alpha x_1 + \beta x_2) = \alpha B(x_1) + \beta B(x_2)$ i.e. the function $B$ is linear.
Now, recall that a linear operator on a normed linear space is bounded iff it is continuous. Given $x_1, x_2 \in X$, then
\begin{equation}
\| B(x_1) - B(x_2) \|
\leq \| B(x_1) - B_k(x_1) \| + \| B_k(x_1) - B_k(x_2) \| + \| B_k(x_2) - B(x_2) \|.
\end{equation}
But, $\| B(x_1) - B_k(x_1) \| \to 0$ and $\| B(x_2) - B_k(x_2) \| \to 0$ as $k \to \infty$, so it remains only to consider the middle term. For each $k$, $B_k$ is bounded and since a Cauchy sequence is bounded, there exists $M > 0$ such that $\| B_k \| \leq M$ for all $k$. It follows that
\begin{equation}
\| B_k(x_1) - B_k(x_2) \| \leq \| B_k \| \| x_1 - x_2 \| \leq M \| x_1 - x_2 \|.
\end{equation}
Therefore, replacing the above inequality into our original inequality and then taking the limit as $k \to \infty$, we find that
$$
\| B(x_1) - B(x_2) \| \leq M \| x_1 - x_2 \|,
$$
that is, $B$ is a bounded operator. We conclude that $B \in B(X,Y)$ and it is a limit of the Cauchy sequence $\{ B_k \}_{k=1}^{\infty}$. Therefore, $B(X,Y)$ is complete in the operator norm, and is thus itself a Banach space.
Best Answer
You may want to use the principle of uniform boundedness.
Since the limit of $T_n(x)$ exists for all $x\in X$, $\|T_n(x)\|\le C_x$ must be bounded for all $x\in X$ (note that the bound depends on $x$, but not on $n$). The PUB now states that $\|T_n\|$ is bounded.
As Davide Giraudo already noted, the PUB is a consequence of Baires category theorem.