Consider the sequence of functions
\begin{equation} f_n(x) = x^n, \quad x \in [0, 1]. \end{equation}
Is this sequence Cauchy in $(C[0, 1], \|\cdot\|_{\infty})$?
The pointwise limit is not continuous because we have
\begin{equation} f(x) := \lim_{n \to \infty} f_n(x) = \begin{cases} 0 & \mbox{if $x \in [0, 1)$} \\ 1 & \mbox{if $x = 1$} \end{cases} \end{equation}
Moreover we have
\begin{align} \|x^n - x^m\|_{\infty} &= \sup_{x \in [0, 1]} |x^n - x^m| \\ &\leq \sup_{x \in (0, 1)} |x^n| + |x^m| \\ &\leq 2 \varepsilon. \end{align}
The first inequality uses the triangle inequality and the fact that $x^n - x^m = 0$ for $x \in \{0, 1\}$. The second inequality uses the fact that for $x^n \to 0$ for $x \in (0, 1)$ so we can find an $N$ such that $x^n < \varepsilon$ for any $n > N$.
Thus $f_n$ is Cauchy in $(C[0, 1], \|\cdot\|_{\infty})$, which is complete. Hence $f_n \to f \in (C[0, 1], \|\cdot\|_{\infty})$. But we already showed that $f$ is not continuous. Contradiction.
Can anyone tell me where things are going wrong?