Let $H$ be a Hilbert space over complex numbers and let $U \subset H$.
Then, the closed linear span of $U$ is defined as $$T = \left\{ \sum_{u \in U} c_n u \mid c_n \text{ complex } , \sum_{u \in U} c_n u \text{ converges unconditionally} \right\}.$$
But, how do I see that this set is closed?
It isn't necessarily closed. For example consider $H=\ell^2(\mathbb N)$ and $U=\{e_1-e_2,e_2-e_3,...\}$ (denote from now on $e_k-e_{k+1}$ by $u_k$).
You have that $e_1$ lies in the closure of the linear span, since it is the limit $$e_1=\lim_{N\to\infty}\sum_{n=1}^N\frac{N-n+1}N\ u_n$$ but not in the "closed linear span" as it is defined here. To see that suppose $e_1=\sum_n c_n u_n$ where the sum converges (a weaker condition than unconditional convergence!), from continuity of the scalar product you get (with $k\neq1$): $$1=\langle e_1,\sum_n c_n\,u_n\rangle=\sum_nc_n\langle e_1,u_n\rangle=c_1\\ 0=\langle e_k,\sum_nc_n u_n\rangle=\sum_nc_n\langle e_k,u_n\rangle=c_{k}-c_{k-1}$$ It follows from induction $c_k=1$ for all $k$. So: $$\|e_1-\sum_n^N u_n\|=\|-e_{N+1}\|=1$$ and the sum doesn't converge to $e_1$!