A sequence $\{x_k\}$ in a Banach space $X$ is a Schauder basis of $X$ if every element $x\in X$ has a unique representation $$ x = \sum_{k=1}^\infty c_k x_k $$ with the series converging in the norm of $X$. A basis $\{x_k\}$ is an unconditional basis if the series $\sum_{k=1}^\infty d_k x_k$ converges for all sequences $\{d_k\}$ satisfying $|d_k|\leq |c_k|$ for each $k$.
QUESTION:
If $\{x_k\}$ is an unconditional basis, the Banach-Steinhaus theorem implies that there is a constant $M$ such that $$ \left\|\sum_{k=1}^\infty d_k x_k\right\|\leq M \left\|\sum_{k=1}^\infty c_k x_k\right\| $$ whenever $|d_k|\leq |c_k|$ for all $k$. How is it possible to show it?
One attempt that doesn't use the Banach-Steinhaus theorem. I'm still wandering why the author mentioned the theorem.
1) There are $\{\lambda_k\}_{k=1}^\infty$ such that $|\lambda_k|\leq\infty$ and $$ \sum_{k=1}^\infty d_k x_k = \sum_{k=1}^\infty \lambda_k c_k x_k $$
2) We could define a linear operator $T_\lambda$ and it's bounded by the definition of the unconditional basis $$ T_\lambda x = \sum_{k=1}^\infty \lambda_k c_k x_k < \infty $$
3) By the definition of a bounded operator $\|T_\lambda x\| \leq M\|x\| $
4) The result follows by observing that $x =\sum_{k=1}^\infty c_k x_k$ .