Show that a subspace of $\ell^2$ is not complete I would like to know if this exercise is correct.
Let $\Bbb R^\infty=\{x:\Bbb N\rightarrow \Bbb R: \exists n \text{ such that}\quad x(k)=0 \quad \forall k\geq n\}$. Show that $(\Bbb R^\infty, \| \cdot\|_{l^2})$ is not complete.
Let $x\in l^2$ be such that $x(k)= ({1\over 2})^k$. Let $(x_n)_n$ a sequence in $\Bbb R^\infty$ defined as: $$x_n(k)= \begin{cases} ({1\over 2})^k, & \mbox{if } k\leq n \\ 0 & \mbox{else }\end{cases}.$$
If we show that $x_n\rightarrow x$ in $l^2$ we've done, since $x\notin \Bbb R^\infty$.
So, $$x(k)-x_n(k)= \begin{cases} 0, & \mbox{if } k\leq n \\({1\over 2})^k  & \mbox{if }k\geq n+1 \end{cases}$$ for each $n\in \Bbb N$.
Now, $\|x-x_n\|_{l^2}^2=\sum_{k=1}^\infty|x(k)-x_n(k)|^2=\sum_{k=n+1}^\infty ({1\over 2})^{2k} $, for each $n\in \Bbb N$. 
Thus $\lim_{n\to \infty} \|x-x_n\|_{l^2}^2=\lim_{n \to \infty}\sum_{k=n+1}^\infty ({1\over 2})^{2k}=0.$ Since $x_n\rightarrow x$ in $l^2$, $(\Bbb R^\infty, \| \cdot\|_{l^2})$ is not complete.

Your proof is correct. But with almost no extra work you can prove a better result: considered as a subset of $\ell^2$, the space $\mathbb R^\infty$ is dense. (Consequently, not closed, hence not complete in the $l^2$ metric.)
The proof goes just as yours, but without specifying $x$. Just let $x$ be any element of $l^2$, then define 
$$x_n(k)= \begin{cases} x(k)  & \mbox{if } k\leq n \\ 0 & \mbox{else }\end{cases}$$
Observe that 
$$
\|x-x_n\|^2 = \sum_{k>n} |x(k)|^2
$$
which converges to zero, being the tail of a convergent series. 
