Am trying to prove if $B$ is a space of all linear maps from $X$ to $ \Omega $ s.t. $\|T\| < \infty$, where $\|T\|=\sup\{\|Tx\|_\Omega : \|x\|_X \leq 1\}$, such that $\Omega$ is complete, then $B$ is complete.
SO far I tried the following sketch:-
let $\sum_n T_n$ s.t. $\sum_n \|T_n\| < \infty $ as $n \to \infty$
\begin{align} \implies & \exists Tx \in \text{s.t. } \|Tx - \sum_nT_nx\| < \varepsilon \text{ (since } \Omega \text{ is complete)} \\[8pt] \implies & \sum_nT_nx -\varepsilon < T_n < \sum_n T_nx + \varepsilon, \text{ as } n \to \infty, \forall x \in X \text{ s.t. } \|x\| \leq 1 \\[8pt] \implies & \|T_x\| < \infty \\[8pt] \implies & T \in B(X,\Omega) \\[8pt] \implies & \sum_nT_n \text{ s.t. } T_n \in B(X,\Omega) \end{align} and $\sum_n \|T_n\| < \infty \implies \exists T \in B(X,\Omega) \text{ s.t. } \|T-\sum_nT_n\| < \varepsilon$ as $n \to \infty \implies B(X,\Omega)$ is complete.
The first arrow, you should denote the norm with $\|\cdot\|_\Omega$, maybe what you want to say is: for each fixed $x\in X$, $$\sum_{n=1}^\infty \|T_nx\|_\Omega \leq \sum_{n=1}^\infty \|T_n\|_{op}\|x\|_X < \infty$$ since $\Omega$ is complete, (by the lemma you mentioned) there exists an element in $\Omega$, denote as $Tx$ such that the partial sum $\sum_{n=1}^k T_nx \rightarrow Tx$ under the norm in $\Omega$ as $k\rightarrow \infty$.
The second arrow, I think there is a typo between the inequality, should not be $T_n$. And more importantly, note that $\sum_{n=1}^k T_n x$ is an element in $\Omega$, and I assume your $\epsilon$ is an element in $\mathbb{R}$, you can not just subtract two elements in different vector spaces. You need to put $\|\cdot\|_\Omega$ around $\sum_{n=1}^k T_n x$.
After the first arrow, you just need to show that the $T$ defined by $$T(x) := \lim_k \sum_{n=1}^k T_n(x)$$is a bounded linear operator from $X$ to $\Omega$.