It is the Exercise 1.1 in Topics in Banach Space Theory by Albiac and Kalton to prove Mazur's Weak Basis Theorem, which states that every weak basis in a Banach space $X$ is a Schauder basis, where weak basis is defined as follows:

A sequence $(e_n)_n$ in a Banach space $X$ is called a weak basis for $X$ if for every $x\in X$ there is an unique sequence of scalars $(a_n)$ such that $x=\sum_{k=1}^\infty a_ke_k$ in the weak topology.

I tried to show, to no avail, that $\lVert\sum^n a_kx_k\rVert\rightarrow\lVert x\rVert$ so that it would immediately follow that $\sum^n a_ne_n\overset{\lVert\cdot\rVert}{\rightarrow}x$. I'd like suggestions on how to proceed.

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Define the closure of the linear hull of the $(x_k)$ in the strong topology by $$ C = \operatorname{cl \ span}\{x_k, k\in\mathbb N\}. $$ It is convex and closed. We have to show that $C=X$ under the condition of weak basis. Let $x\in X\setminus C$ be given. Then we can separate $x$ and $C$ by $f\in X^*$: There is $\epsilon>0$ such that $$ f(x) + \epsilon \le f(y) \quad\forall y\in C. $$ Since $C$ is a linear space, $f(y)=0$ for all $y\in C$. In particular, $f(x_k)=0$ for all $k$. By assumption, there is a sequence $(a_k)$ such that $\sum_{k=1}^n a_kx_k$ converges weakly to $x$ for $n\to\infty$. This implies $$ f(x) = \lim_{n\to\infty} f(\sum_{k=1}^n a_kx_k) = \lim_{n\to\infty} \sum _{k=1}^n a_k f(x_k)=0, $$ which is a contradiction to $f(x)+\epsilon<0$.

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