Prove that $\overline{\mathrm{span}\lbrace x_k\rbrace_{k=1}^\infty}=\ell^2$ where $x_k := (0, ..., \underset{\text{k-th spot}}{1/\sqrt k}, 0, 0,...,)$

Question

Define $$ x_k := (0, ..., \underset{\text{k-th spot}}{1/\sqrt k}, 0, 0,...,), $$ for all $k, n \in \mathbb{N}$, then indeed $$ \overline{\mathrm{span}\lbrace x_k \rbrace_{k=1}^\infty} = \ell^2. $$ I would like to check this statement. Then I thought to do as follows. First, the set $x_k$ is clearly orthogonal. Then for each $x\in\ell^2$ we have $$x=\sum_k \langle x,x_k\rangle x_k$$ and $$\|x\|^2=\sum_k |\langle x,x_k\rangle|^2$$ Orthogonal decomposition of $\ell^2$: $\ell^2=X\oplus X^{\perp}$ where $X=\overline{\mathrm{span}\lbrace x_k \rbrace_{k=1}^\infty}$. If I consider $x\in X^{\perp}$ we have that $\langle x, x_k\rangle=0$ for each $k$. Thus $\|x\|^2=\sum_k |\langle x,x_k\rangle|^2=0$ and so $x=0$. But this means that $X^{\perp}=\{0\}$ and thus the claim. Right?