Let $$f_k(t) = \begin{cases}0, \text{ if } t \in [0,1/2]\\ 1, \text{ if } t \in [1/2 + 1/k, 1].\end{cases}$$We will prove the sequence $(f_k)_{k \in \mathbb{N}}$ is Cauchy on $(C[0,1],\|\cdot\|_p)$ for $1 \leq p < \infty$. (this can also be done for $\infty$ but for the sake of the exercise we shall omit that case). Notice that \begin{align} \|f_n-f_n\|_p & = \left(\int_0^1|f_n(t)-f_m(t)|^pdt\right)^{1/p}\\ & \leq 2\left(\int_{1/2+1/n}^1|f_n(t)|^pdt - \int_{1/2 + 1/m}^1|f_m(t)|^pdt\right)^{1/p}\\ & \leq 2\left(\int_{1/2+1/n}^1dt - \int_{1/2+1/m}^1dt\right)^{1/p}\\ & \leq 2\left(1-1/2-1/n - 1 + 1/2 + 1/m\right)^{1/p}\\ & = 2\left(\frac{1}{m} - \frac{1}{n}\right)^{1/p}. \end{align}And when $n,m \rightarrow \infty$ we have $\|f_n-f_m\| \rightarrow 0$. Rigorously, we could pick $N > 0$ so that when choosing $n>m\geq N$ we have $1/m-1/n < \epsilon$.
Can anyone verify if my reasoning is correct? Thanks in advance!!
As discussed in the comments, the functions $f_k$ are continuous and somewhere in $[0,1]$ on $[1/2,1/2+1/k]$. Then we have \begin{align*} \|f_n-f_m\|_p^p=\int_{1/2}^{1/2+1/N}|f_n(x)-f_m(x)|^p\mathrm dx \le\int_{1/2}^{1/2+1/N}1^p\mathrm dx =\frac{1}{N} \end{align*} for $n\ge m\ge N$, where we used that $|f_n(x)-f_m(x)|\le 1$ since $f_n(x),f_m(x)\in[0,1]$, and that $x^p$ is increasing for $x\ge 0$. So, for any $\varepsilon\in(0,1)$ we take $N=\varepsilon^{-p}$, then we get $\|f_n-f_m\|_p\le\varepsilon$ for all $n,m\ge N$, so the sequence is Cauchy.