Consider a sequence of objects (i.e. functions in this case) $(f_{n}) \in B_{E}(X)$. Let this sequence be cauchy in $B_{E}(X)$.
Now we need to find a candidate function in $B_{E}(X)$ that would satisfy our needs. So lets change the space to E and consider $[f_{n}(x)]$ where $x\in X$ (a sequence of objects in E). Now for a fixed $x \in X$. This is still a cauchy sequence in E and E is Banach. Hence the sequence would converge to an object in E. Hence as we change $x\in X$, we get a new converged object in E for every $x\in X$ . This can now be called a mapping from X to E. Lets represent it by $F:X\rightarrow E$. $F(x)$ is thus our candidate function.
Now you need to prove two more things
a. $f_{n}\rightarrow F$ (Why? Because $f_{n}(x) \in E$ while $f_{n} \in B_{E}(X)$. In other words, we have only proven pointwise convergence in E but now we have prove convergence in $B_{E}(X)$)
b. prove that $F$ is bounded implying $F\in B_{E}(X)$
Now lets prove (a). Remember $(f_{n}) \in B_{E}(X)$ is cauchy in $B_{E}(X)$. Hence, $\forall m,n \geq N \|f_{n}-f_{m}\|<\epsilon$. Thus, for an arbitrary $x\in X$ $\|f_{n}(x)-f_{m}(x)\|_{E}<\epsilon$. With $n$ fixed and $m \rightarrow \infty$ $\|f_{n}(x)-F(x)\|_{E}<\epsilon$. Since $x\in X$ was arbitrary, then its true $\forall x \in X$. Hence $\|f_{n}-F\| = \sup_{x\in X}\|f_{n}(x)-F(x)\|_{E}<\epsilon$ and $f_{n}\rightarrow F$ in $B_{E}(X)$.
You see its rather confusing to switch between spaces and using appropriate norms for them, otherwise these proofs are quite standard.
Use the proof for $E = \mathbb{R}$ to prove boundedness.
Take $\sqrt{x+1/n}$, which is $C^1[0,1]$ for each $n$ thanks to our shifting of the discontinuity in the derivative.
To see this converges uniformly, note that $\sqrt{x}$ is uniformly continuous on compact sets. On say $[0,2]$ we may use uniform continuity to meet any epsilon challenge with an $N$ not dependent on $x$ with
$$
|\sqrt{x+1/N}-\sqrt{x}|<\epsilon
$$
Best Answer
As I pointed out in the comments, this follows directly since you are dealing with a finite-dimensional normed space over a complete field (namely $\mathbb{C}$). However, if you want to prove it directly, you do it as follows.
Let $\{A^{(k)}\}_{k=1}^\infty$ be Cauchy, i.e. $$ \Vert A^{(n)} - A^{(m)} \Vert < \epsilon $$ when $n,m \geq N$ for some $N$. This is equivalent to saying $$ |a^{(n)}_{jk}-a^{(m)}_{jk}| \leq \max_k \sum_{j=1}^3 |a^{(n)}_{jk}-a^{(m)}_{jk}| < \epsilon, $$ whenever $n,m > N$, so that $\{a^{(n)}_{jk}\}_{n=1}^\infty$ is Cauchy in $\mathbb{C}$ for each $j,k$, and since $\mathbb{C}$ is complete, it converges, i.e. $a^{(n)}_{jk}\to \tilde a_{jk} \in \mathbb{C}$ as $n\to \infty$. Put $$ \tilde A = \begin{pmatrix}0 & \tilde a_{12} & \tilde a_{13}\\0 & 0 & \tilde a_{23}\\ 0&0&0\end{pmatrix}. $$ Then $\tilde A$ clearly belongs to your space. Can you now prove that $A^{(n)} \to \tilde A$ as $n \to \infty$ to finish the proof?