To show that $(C_b(X), \| \cdot \|_\infty)$ is complete we first show that there is a pointwise limit function in $\mathbb{R}$ to which $f_n$ converges. For this we note that because $f_n$ is Cauchy with respect to the sup norm, it follows that $f_n(x)$ is a Cauchy sequence in $\mathbb{R}$ for any $x$ in $X$. But $\mathbb{R}$ is complete and hence the limit $\lim_{n \to \infty} f_n (x)$ is in $\mathbb{R}$.
Now let $f(x)$ denote the pointwise limit function of $f_n$. We now want to show that $f$ is bounded, that is, there exists a real constant $K$ such that $\| f \|_\infty < K$. For this we again use that $f_n$ is Cauchy with respect to the sup norm: For every $\varepsilon > 0$ we can find an $N$ such that for $n,m \geq N$ we have that $\| f_n - f_m \|_\infty < \varepsilon$. Using the triangle inequality we have $\| f \|_\infty \leq \| f - f_N \|_\infty + \| f_N \|_\infty$ and because $f_N$ is in $C_b(X)$ we know that there exists an $M$ in $\mathbb{R}$ sucht that $\| f_N \|_\infty \leq M$. We also have $\| f_n - f_N \|_\infty < \varepsilon$ for all $n \geq N$ and hence $\lim_{n \to \infty} \| f_n - f_N \|_\infty = \| f - f_N \|_\infty \leq \varepsilon$. Hence $f$ is bounded.
Now we want to show that $f_n$ converges to $f$ in norm, that is, $\| f - f_n \|_\infty \xrightarrow{n \to \infty} 0$. For this let $\varepsilon > 0$. Then we have that there exists an $N$ such that for $n,m \geq N$, $\| f_n - f_m \|_\infty < \frac{\varepsilon}{2}$, again because $f_n$ is Cauchy.
Using the triangle inequality we get $\| f - f_n \|_\infty \leq \| f - f_N \|_\infty + \| f_N - f_n \|_\infty \leq \varepsilon$. By the same argument as before, that is, because $f_n$ is Cauchy, $\| f_n - f_N \|_\infty < \frac{\varepsilon}{2}$ for all $n \geq N$ and hence $\lim_{n \to \infty} \| f_n - f_N \|_\infty = \| f - f_N \|_\infty \leq \frac{\varepsilon}{2}$.
So $\| f - f_n \|_\infty \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$ and as $\varepsilon$ was arbitrary it follows by having $\varepsilon$ tend to $0$ that $\| f - f_n \|_\infty \xrightarrow{n \to \infty} 0$.
Finally, now that we have convergence in norm, we can apply the uniform limit theorem to get that $f$ is continuous and hence $f$ is in $C_b(X)$.