I am supposed to prove that the set of all bounded real functions on a metric space $\displaystyle X$ with metric $\displaystyle d(x,y)=\left\|f-g\right\|_{\infty}=\sup_{x\in X}\left|f(x)-g(x)\right|$ is complete.

So we have some Cauchy sequence $\displaystyle \left\{f_n\right\}_{n\in\mathbb{N}}$. If we fix $\displaystyle \xi\in X$ we can clearly see that $\displaystyle \left\{f_n(\xi)\right\}_{n\in\mathbb{N}}$ is a Cauchy sequence in $\displaystyle \mathbb{R}$ and thus there exists some $\displaystyle \alpha\in\mathbb{R}$ such that $\displaystyle f_n(\xi)\to\alpha$. So, define $\displaystyle \bar{f}:X\mapsto\mathbb{R}$ by $\displaystyle \bar{f}(\xi)=\lim\text{ }f_n(\xi)$. It is an elementary concept from analysis that a Cauchy sequence of functions converges uniformly. Thus, we can make $\displaystyle \left|f_n(x)-\bar{f}(x)\right|<\varepsilon$ by making $\displaystyle n$ sufficiently large regardless of what value $\displaystyle x$ is. Thus, for every $\displaystyle \varepsilon>0$ there exists some $\displaystyle N\in\mathbb{N}$ such that $\displaystyle n\leqslant N\implies \left|f_n(x)-\bar{f}(x)\right|<\varepsilon,\text{ }\forall x\in X$ and so $\displaystyle n\leqslant N\implies \left\|f_n(x)-\bar{f}(x)\right\|_{\infty}<\varepsilon$. So, it remains to show that $\displaystyle \bar{f}(x)$ is bounded. But, this is trivial, since there exists some $\displaystyle n\in\mathbb{N}$ such that $\displaystyle \left|\bar{f}(x)\right|-\left|f_n(x)\right|\leqslant\left|\bar{f}(x)-f_n(x)\right|\leqslant 1$ for every $\displaystyle x\in X$ and thus $\displaystyle \left|\bar{f}(x)\right|\leqslant\left|f_n(x)\right |+1$ but since $\displaystyle f_n(x)$ is bounded for every $\displaystyle n$ we see that there exists some $\displaystyle M\in\mathbb{R}$ such that $\displaystyle \left|\bar{f}(x)\right|\leqslant M+1$ and thus $\displaystyle \bar{f}(x)$ is bounded and thus it is in our metric space.

So, we may conclude that this space is complete.

How's that look?