Let

$$f_{n,m}(x):=\begin{cases}n (1- m x) & 0?x?\frac1m \\ 0 & \text{else}\end{cases}$$

this is continuous function on $[0,1]$. For $p<\infty$ its $L^p$ norm is

$$\|f_{n,m}\|_p = \int_0^1|f(x)|^p \,dx = n^p\int_0^{1/m} (1-mx)^p\,dx = \frac{n^p}m \int_0^1 (1-x)^p\,dx = \frac{n^p}{m\cdot (p+1)}$$

and the $L^p$ norm for $\|(f_{n,m})^2\|_p$ is:

$$\|(f_{n,m})^2\|_p = ... = \frac{n^{2p}}{m\cdot (2p+1)}$$

It follows that

$$\|f_{n,n^{3p/2}}\|_p = \frac{1}{n^{p/2}(p+1)},\qquad\|(f_{n,n^{3p/2}})^2\|_p=\frac{n^{p/2}}{(2p+1)}$$

so $f_{n,n^{3p/2}}\to0$ as $n\to0$ in $L^p$-norm, but $(f_{n,n^{3p/2}})^2$ does not converge to $0$ in this norm - hence squaring is not continuous on $L^p$ for $p<\infty$.

On the other hand if $f_n\to f$ in $L^\infty$, both $f_n$ and $f$ continuous, you have that:

$$\|f_n^2-f^2\|_\infty = \|(f_n-f)\cdot (f_n+f)\|_\infty ?\|f_n-f\|_\infty \cdot \|f_n+f\|_\infty? \|f_n-f\|_\infty \cdot (\|f_n\|_\infty + \|f\|_\infty)$$

now since $f_n\to f$ in $L^\infty$ you have that

$$\|f_n\|_\infty = \|f+ f_n-f\|_\infty ? \|f\|_\infty + \|f_n - f\|_\infty$$

whence it follows that there is some constant $D$ so that $\|f_n\|_\infty<D$ for all $n$. So we have showed that:

$$\|f_n^2-f^2\|_\infty ? \|f_n-f\|_\infty \cdot (D+\|f\|)$$

whence also $\|f_n^2-f^2\|_\infty\to 0$, meaning nothing else than $f_n^2\to f^2$ in $L^\infty$. This verifies that squaring is sequentially continuous on $(C[0,1], d_\infty)$ - hence continuous.