Show that $l^2$ is a Hilbert space 
Let $l^2$ be the space of square summable sequences with the inner product $\langle x,y\rangle=\sum_\limits{i=1}^\infty x_iy_i$.
  (a) show that $l^2$ is H Hilbert space. 

To show that it's a Hilbert space I need to show that the space is complete. For that I need to construct a Cauchy sequence and show it converges with respect to the norm. However, I find it confusing to construct a Cauchy sequence of sequences?

A typical proof of the completeness of $\ell^2$ consists of two parts.
Reduction to series
Claim: Suppose $ X$ is a normed space in which every absolutely convergent series converges; that is, $ \sum_{n=1}^{\infty} y_n$ converges whenever $ y_n\in X$ are such that $ \sum_{n=1}^{\infty} \|y_n\|$ converges. Then the space $X$ is complete.
Proof. Take a Cauchy sequence $ \{x_n\}$ in $X$. For $ j=1,2,\dots$ find an integer $ n_j$ such that $ \|x_n-x_m\|<2^{-j}$ as long as $ n,m\ge n_j$. (This is possible because the sequence is Cauchy.) Also let $ n_0=1$ and consider the series $ \sum_{j=1}^{\infty} (x_{n_{j}}-x_{n_{j-1}})$. This series converges absolutely, by comparison with $\sum 2^{-j}$. Hence it converges. Its partial sums simplify (telescope) to $ x_{n_j}-x_1$. It follows that the subsequence $ \{x_{n_j}\}$ has a limit. It remains to apply a general theorem about metric spaces: if a Cauchy sequence has a convergent subsequence, then the entire sequence converges. $\quad \Box$
Convergence of absolutely convergent series in $\ell^2$
Claim:: Every absolutely convergent series in $ \ell^2$ converges
Proof. The elements of $ \ell^2$ are functions from $ \mathbb N$ to $ \mathbb C$, so let's write them as such: $ f_j: \mathbb N\to \mathbb C$. (This avoids confusion of indices.) Suppose the series $ \sum_{j=1}^{\infty} \|f_j\|$ converges. Then for any $ n$ the series $ \sum_{j=1}^{\infty} f_j(n)$ converges, by virtue of comparison $|f_j(n)| \le \|f_j\|$.
Let $ f(n) = \sum_{j=1}^{\infty} f_j(n)$. So far the convergence is only pointwise, so we are not done. We still have to show that the series converges in $ \ell^2$, that is, its tails have small $ \ell^2$ norm: $ \sum_{n=1}^\infty |\sum_{j=k}^{\infty} f_j(n)|^2 \to 0$ as $ k\to\infty$.
What we need now is a dominating function/sequence (sequences are just functions with domain $\mathbb{N}$), in order to apply the Dominated Convergence Theorem. Namely, we need a function $ g: \mathbb N\to [0,\infty)$ such that
$$ \sum_{n=1}^{\infty} g(n)^2<\infty \tag{1}$$
$$ \left|\sum_{j=k}^{\infty} f_j(n)\right|  \le g(n) \quad \text{for all } \ k,n \tag{2} $$
Set $ g(n) =  \sum_{j=1}^{\infty} |f_j(n)| $. Then (2) follows from the triangle inequality. Also, $ g$ is the increasing limit of functions $ g_k(n) = \sum_{j=1}^k |f_j(n)| $. For each $k$, using the triangle inequality in $ \ell^2$, we get
$$ \sum_n g_k(n)^2 = \|g_k\|^2\le \left(\sum_{j=1}^k \|f_j\|\right)^2 \le   S^2$$
where $S= \sum_{j=1}^\infty\|f_j\|$. Therefore, $ \sum_n g(n)^2\le S^2$ by the Monotone Convergence Theorem.
To summarize: we have shown the existence of a summable function $g^2$ that dominates the square of any tail of the series $\sum f_j$. This together with the pointwise convergence of said series yield its convergence in $\ell^2$. $\quad\Box$
