I came across this theorem in textbook Analysis I by Amann/Escher
Let $X$ be a nonempty set and $(E, \| \cdot \|)$ a Banach space. Then the space of bounded functions $\mathcal B(X, E)$ endowed with the sup norm $\| \cdot \|_\infty$ is also a Banach space.
Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
My attempt:
Let $(u_n)$ be a Cauchy sequence in $\mathcal B(X, E)$. Then there exists $N \in \mathbb N$ such that $$\| u_n - u_m \|_\infty < \varepsilon, \quad m,n \ge N$$
or equivalently $$\sup_{x \in X} \| u_n(x) - u_m(x) \| < \varepsilon, \quad m,n \ge N, \quad x\in X$$
Then $$\| u_n(x) - u_m(x) \| \le \sup_{x \in X} \| u_n(x) - u_m(x) \| < \varepsilon, \quad m,n \ge N, \quad x\in X \quad (\star)$$
and thus $(u_n(x))$ is a Cauchy sequence for each $x \in X$. Because $E$ is complete, $(u_n(x))$ is convergent for each $x \in X$. We define $u \in E^X$ by $u(x) := \lim_{n \to \infty} u_n(x)$ for all $x \in X$. Next we prove that $u \in \mathcal B(X, E)$. Take the limit when $m \to \infty$ of $(\star)$, we get $$\| u_n(x) - u(x) \| \le \sup_{x \in X} \| u_n(x) - u(x) \| \le \varepsilon, \quad n \ge N, \quad x\in X \quad (\star \star)$$
Clearly, $u$ is bounded because, by triangle inequality, we have $$\|u(x)\| \le \| u_{N}(x) \| + \| u_N(x) - u(x) \| < u_N(x) + \varepsilon \le \sup_{x \in X} u_N (x) + \varepsilon, \quad x \in X$$
It follows from $(\star \star)$ that $$\| u_n - u \|_\infty = \sup_{x \in X} \| u_n(x) - u(x) \| \le \varepsilon, \quad n \ge N$$
As such, $(u_n)$ converges to $u$. This completes the proof.