A vector basis of a vector space $E$ is a family $(a_{\lambda})_{\lambda\in L}$ such that any element of $E$ can be written in a unique way as a linear combination of a finite number of $a_{\lambda}$, this implies in particular that the $a_\lambda$ are linearly independent.
Let $(a_n)$ be a sequence of linearly independent elements in a Banach space $E$. Define inductively a sequence $(t_n)$ of positive real numbers in the following way: if $d_n$ is the distance of the point $t_n a_n$ to the subspace $V_{n-1}$ generated by $a_1,...., a_{n-1}$ (note that $d_n>0$), take $t_{n+1}$ such that $|t_{n+1}|\cdot ||a_{n+1}|| < \frac{d_{n}}{3}$. Show that the series $\displaystyle{\sum_{n=1}^{\infty}t_n a_n}$ is absolutely convergent, and that its sum $x$ does not belong to any of the subspaces $V_n$.
I assume that $V_0$ is defined to be $\{0\}$. Since $0\in V_{n-1}$, we have $d_n\leq \Vert t_na_n\Vert$, so $\Vert t_{n+1} a_{n+1}\Vert<\frac13 \Vert t_na_n\Vert$ for all $n\in\mathbb N$, by the choice of $t_{n+1}$. It follows (ratio test) that the series $\sum_{n\geq 1} t_n a_n$ is absolutely convergent.
Now let $x=\sum_{k=1}^\infty t_ka_k$ and, towards a contradiction, assume that $x\in V_n$ for some $n$. Put $$z=\sum_{k\geq n+2} t_ka_k\, .$$ Then $z=y-t_{n+1}a_{n+1}$ for some $y\in V_{n}$, namely $y=x-\sum_{k\leq n} t_ka_k$. By the choice of $t_{n+2}$, we have $\Vert t_{n+2}a_{n+2}\Vert<\frac 13 \Vert t_{n+1}a_{n+1}-y\Vert=\frac 13\Vert z\Vert$. Moreover, as observed above we also have $\Vert t_{i+1}a_{i+1}\Vert<\frac13 \Vert t_ia_i\Vert$ for any $i\in\mathbb N$, so that $\Vert t_{k}a_k\Vert< \frac{1}{3^{k-n-2}} \Vert t_{n+2}a_{n+2}\Vert$ for all $k\geq n+2$. Hence, we get $\Vert t_{k}a_k\Vert<\frac1{3^{k-n-1}}\Vert z\Vert$ for all $k\geq n+2$. On the other hand, $\Vert z\Vert\leq \sum_{k\geq n+2} \Vert t_ka_k\Vert$ by the triangle inequality, so we obtain $$\Vert z\Vert<\sum_{k\geq n+2} \frac1{3^{k-n-1}}\Vert z\Vert=\left(\frac13+\frac1{3^2}+\cdots \right)\Vert z\Vert=\frac12\Vert z\Vert\, . $$ Because of the strict inequality, this is a contradiction.