Prove that the space $C^k(\bar{U})$ is Banach. Definition: Let $U\subset \mathbb{R}^n$ be an open. Define the space,
$C^k(\bar{U})=\{u\in C^k(U):D^{\alpha}u 
 \text{ is bounded and uniformly continuous } \forall \alpha:|\alpha|\leq k\}$
with the norm, 
$||u||_{C^k(\bar{U})}=\sum_{\alpha:|\alpha|\leq k}\underset{{x\in U}}{\sup}|D^{\alpha}u(x)|$. 
Proposition: $\Big(C^k(\bar{U}),||\cdot||_{C^k(\bar{U})}\Big)$ is a Banach space.
My attempt at a proof, 
Proof (Unfinished). The positive definiteness, homogeneity and triangle inequality follow from a simple calculation. 
We need to show that every Cauchy sequence converges in $C^k(\bar{U})$. Let $(u_k)_{k\in \mathbb{N}}\subset C^k(\bar{U})$ be Cauchy, this means that, 
$(\forall \epsilon>0) (\exists N\in\mathbb{N})$ such that if $n,m\geq N$, $||u_m-u_n||_{C^k(U)}<\epsilon$. 
This implies, 
$\sum_{\alpha:|\alpha|\leq k}\underset{{x\in U}}{\sup}|D^{\alpha}u_m(x)-D^{\alpha}u_n(x)|<\epsilon$
and therefore 
$\underset{{x\in U}}{\sup}|D^{\alpha}u_m(x)-D^{\alpha}u_n(x)|<\epsilon\implies|D^{\alpha}u_m(x)-D^{\alpha}u_n(x)|<\epsilon$ 
Hence $(D^{\alpha}u_n(x))_{n\in\mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$ $\forall x\in U$. $\mathbb{R}$ is complete and hence every Cauchy sequence converges in $\mathbb{R}$. Denote the limit in the $\alpha=(0,...,0)$ as $u$. For the cases where $\alpha\neq (0,...,0)$ denote the limit as $g^{\alpha}(x)$. 
At this point I get stuck, I think at this point I am left to argue the following:
1.Uniform convergence of $u_m\rightarrow u$.
2.$g^{\alpha}(x)= D^{\alpha}u(x)$ and $D^{\alpha}u_m\rightarrow D^{\alpha}u$ uniformly.


*$u\in C^{k}(\bar{U})$
Any help would be greatly appreciated. 
 A: You are on the right lines. Once you have got that $\sup_{x\in U}|D^\alpha u_n - D^\alpha u_m| < \epsilon$ for each $|\alpha|\leq k$, this tells you that the sequence $(D^\alpha u_n)_n$ is a uniformly Cauchy sequence. You can show (quite easily) that any uniformly Cauchy sequence actually converges uniformly, and so instead of the pointwise convergence which you had, we actually have $D^\alpha u_n\to g^\alpha$ uniformly. 
Next note that boundedness and uniform continuity are inherited by uniform limits, and thus we know all the $g^\alpha$ are bounded and uniformly continuous.
To finish we just need to show (if $u = g^0$) that $D^\alpha u = g^\alpha$, which would answer your third question. Then we will have $D^\alpha u_n\to D^\alpha u$ uniformly for all $|\alpha|\leq k$, which shows $u_n\to u$ in $C^k(\overline{U})$ (answering your second question). Deducing that $D^\alpha u = g^\alpha$ simply involves arguing that we exchange the uniform limit with the derivative, since this reads $$D^\alpha (\lim_{n\to\infty} u_n) = \lim_{n\to \infty} D^\alpha u_n.$$ This can be shown inductively, by noting that in the real case, if $(f_n)_n$ is a sequence of $C^1$ functions on an interval, and $f_n\to f$ uniformly and $f_n’\to g$ uniformly, then in fact $f$ is differentiable and $f’ = g$. This can be shown via the fundamental theorem of calculus.
However you should note that the proof of this later fact, since it relies on the fundamental theorem of calculus, relies on $U$ being path connected (or equivalently connected here) and being of finite diameter. Thus in the case where $U$ is connected of finite diameter, the result is proven. In the case where $U$ is not connected, we can just apply the result on each connected component to get $D^\alpha u = g^\alpha$ on each component, and thus they agree everywhere.
In the case when $U$ does not have finite diameter, but is still connected, the result is true by taking a cover of $U$ by open increasing nested sets $U_n$ of finite diameter, and then apply the above on each $U_n$ (noting that each successive limit must agree with the previous on the overlap). You can then show the existence of a global pointwise limit $u$, and then show that the sequence converges uniformly to $u$ in the same way that being uniformly Cauchy implied uniform convergence.
