Showing $E(\Omega)$ is a Hilbert Space Let $E(\Omega)=\{ u\in \{L^2 (\Omega)\}^n : \text{div } u \in L^2(\Omega)\}$, that is $E(\Omega)$ consists of vector valued functions $u=(u^1, \cdots , u^n)$ where each component function $u^i$, $i\in \{ 1, \cdots n\}$ is a $L^2(\Omega)$ function and $\frac{\partial u^1}{\partial x_1}+\cdots+\frac{\partial u^n}{\partial x_n}\in L^2(\Omega)$.
We want to show that under the norm induced by this inner product
$$
(u,v)_{E(\Omega)}=\sum_{i=1}^{n} (u_j, v_j)_{L^2(\Omega)}+(\text{div } u, \text{div }v)_{L^2(\Omega)}
$$ 
$(E(\Omega), ||.||_{E(\Omega)})$ is a Hilbert Space.
This is what I manage to do. 
Suppose $\{u_m=(u^1, \cdots , u^n)\}_{m\in \mathbb{N}}$ is a Cauchy sequence in $(E(\Omega), ||.||_{E(\Omega)})$.
Fix $i \in \{1,\cdots, n \}$. Because $(u^i_m-u^i_l, u^i_m-u^i_l )$=$||u^i_m-u^i_l ||^2_{L^2(\Omega)} \leq  ||u_m-u_l||^2_{E(\Omega)}$ and the fact that $\{u_m=(u^1, \cdots , u^n)\}_{m\in \mathbb{N}}$ is a Cauchy sequence in $(E(\Omega), ||.||_{E(\Omega)})$, we see that $\{u^i_m\}_{m\in\mathbb{N}}$ is Cauchy in $L^2(\Omega)$. By the completeness of $L^2(\Omega)$, we concluded that $u^i_m \rightarrow u^i \in L^2(\Omega)$ in $L^2$ norm as $m\rightarrow \infty$. Hence we concluded that $u=(u^1, \cdots, u^n) \in \{L^2(\Omega)\}^n$.
Also note that 
$$
(\text{div }(u_m-u_l), \text{div }(u_m-u_l))_{L^2(\Omega)}=||\text{div } u_m - \text{div } u_l||^2_{L^2(\Omega)}\leq ||u_m-u_l||^2_{E(\Omega)}
$$
Hence $\{\text{div } u_m\}_{m\in\mathbb{N}}$ is Cauchy  in $L^2(\Omega)$. Again by the completeness of $L^2(\Omega)$, we can conclude that $\text{div } u_m \rightarrow g\in L^2(\Omega)$ in $L^2$ norm as $m \rightarrow \infty$.
Now to show $u \in E(\Omega)$, we need to show $\text{div } u$ exists and is also in $L^2(\Omega)$. This is achieved by showing $g=\text{div }u$. But this is where I get stuck, how do we show this last part? I don't even know that the partial derivative of $u^i$ exists. 
Thanks. 
 A: 
We need to show $\operatorname{div} u$ exists and is also in $L^2(\Omega)$. This is achieved by showing $g=\operatorname{div}u$. But this is where I get stuck, how do we show this last part? 

As I said in the comments, the divergence of $u$ is actually the weak divergence of $u$. First the definition of the weak derivative for an $L^2$-integrable function $v$:
$$
w := \text{Weak derivative of }v \text{ w.r.t }x_i \iff \int_{\Omega} w \phi = -\int_{\Omega} v \,\partial_{x_i}\phi,\tag{$\star$}
$$
for any smooth function $\phi$ with compact support in $\Omega$ (this is the test function space, we denote it as $C^{\infty}_c(\Omega)$). Now if we say $\partial_{x_i} v$, we actually mean $w$ given by $(\star)$.
The weak divergence is defined in a similar way:
$$
g := \operatorname{div} u \iff \int_{\Omega} g \phi = -\int_{\Omega} u \cdot \nabla \phi, \quad\text{for any } \phi\in C^{\infty}_c(\Omega).\tag{1}
$$
What you wanna show now is just (1) holds. Now we use the definition of weak divergence on $u_m$:
$$
\int_{\Omega} \phi \operatorname{div} u_m  = 
-\int_{\Omega} u_m \cdot \nabla \phi. 
$$
Notice 
$$
\operatorname{div} u_m \to g \;\text{ in }\|\cdot\|_{L^2(\Omega)} \implies \int_{\Omega} \phi \operatorname{div} u_m \to \int_{\Omega} \phi g ,
\\
u_m \to u \;\text{ in }\|\cdot\|_{L^2(\Omega)} \implies \int_{\Omega} u_m\cdot \nabla \phi \to \int_{\Omega} u\cdot \nabla \phi.
$$
Why these two convergence results are true? Cauchy-Schwarz inequality is your friend (left for you). Hence (1) holds, and $\operatorname{div} u = g\in L^2(\Omega)$.


I don't even know that the partial derivative of $u^i$ exists.

Good call. The partial derivative of $u^i$ may not exist pointwisely indeed, it is well-defined in the distribution sense, a.k.a., $(\star)$ holds if we switch $v$ as a component of $u$ (the fact that each component of $u$ is $L^2$-integrable is used as well). 

Backstory: The space $E(\Omega)$ here is the famous $H(\operatorname{div})$ space. Any functional analysis book covering incompressible flow, or Navier-Stokes equations, should have its introduction. The proof of $H(\operatorname{div})$ is Hilbert very similar to the one of $H^1 := W^{1,2}$ so I guess most book would leave this as an exercise. $H(\operatorname{div})$ is the completion of the smooth vector fields in the graph norm (the $H(\operatorname{div})$-norm):
$$
\left(\|\cdot\|_{L^2(\Omega)}^2 + \|\operatorname{div}(\cdot)\|_{L^2(\Omega)}^2\right)^{1/2}.
$$
