Proving that the linear space of all elements of the form $\sum_{n=1}^{\infty}a_{n}x_{n}$ is closed

Question

I want to proof the following result:

1. Let $(X,\left \| \cdot \right \|)$ be a Banach Space. Let $(x_n)_{n\in \mathbb{N}}\subseteq X$ be such that $x_n\neq 0$ for all $n\in \mathbb{N}.$

2. There is a constant $K$ such that, for every choice of scalars $(a_i)_{i=1}^{\infty}$, and integers $n<m$, we have

$$\left \| \sum_{i=1}^{n}a_{i}x_{i} \right \|\leq K\left \| \sum_{i=1}^{m}a_{i}x_{i} \right \|$$

3. The closed linear span of $(x_n)_{n\in \mathbb{N}}$ is all $X$

Then, the set of all elements of the form $\sum_{n=1}^{\infty}a_{n}x_{n}$ is closed.

In fact, this is a characterization for Schauder basis, but I am only interested in prove this closedness. I have tried to prove this by a diagonalization process, and it doesn't work; I also have tried to prove this by proving that the sequence of $(a_{i}^{(n)})_{n\in \mathbb{N}}$ is a cauchy sequence if $(y_{n})_{n\in \mathbb{N}}$ is a convergent sequence of the form

$$\sum_{i=1}^{\infty}a_{i}^{(n)}x_i.$$

I truly think that this last aproximation is the good way. But I still have troubles in the proof. Can someone give a hint in order to prove this result? Any help will be greatly appreciated.