Definition: A sequence in Banach space $X$ is called a Schauder basis for $X$ if for each $x\in X$ there exists a unique sequence $(\alpha_n)$ of scalars such that $x=\lim_n\Sigma_{k=1}^{n}\alpha_k x_k$.
Definition: A sequence $(x_n)$ in Banach space $X$ is called a basic sequence if it is a Schauder basis for its closed linear span $[x_n]$.
I cannot distinguish $[x_n]$ and the space for which the sequence $(x_n)$ form a Schauder basis. I think $(x_n)$ always be a Schauder basis of its closed linear span because every element in $[x_n]$ is a limit of $\Sigma_{k=1}^{n}\alpha_k x_k$ and $\lim_n\Sigma_{k=1}^{n}\alpha_k x_k$ belongs to $[x_n]$.
Where did I wrong? Is there any example?
Well, a first condition is that the $(x_n)$ should be linearly independent, otherwise if e.g. $\sum_{k=1}^n \alpha_k x_k = 0$, then you have two different sequences of coefficients where the linear combination converges to zero: $(\alpha_1, \dots, \alpha_k, 0, 0, \dots)$ and $(0,0,\dots)$.
But even then this is not sufficient. Take for example the Banach space $\ell^2$ of square summable sequences. Let $e^i \in \ell^2$ be given by $e^i_i = 1$ and $e^i_j = 0$ for $j \neq i$. Let also $u \in \ell^2$ be given by $u_n = \frac{1}{n+1}$, and consider the sequence $(u, e^0, e^1, \dots)$.
It is linearly independent, because a linear combination can only involve a finite number of nonzero coefficients. Its closed linear span is all of $\ell^2$. However, it is not a Schauder basis for $\ell^2$, because there are two different sequences of coefficients where the linear combination converges to $u$. The sequence $(1, 0, 0, \dots)$ is the first obvious one: $\lim_{n \to \infty} u = u$. The second sequence is $(0, 1, 1/2, 1/3, 1/4, \dots)$; indeed $$\left\| u - \sum_{k=0}^n \frac{1}{n+1} e^i \right\|_{\ell^2} = \sum_{k=n+1}^\infty \frac{1}{(n+1)^2}$$ which converges to zero as $n \to \infty$.
Basically, you forgot the uniqueness condition in the definition of a Schauder basis.