How to prove that $L^p [0,1]$ isn't induced by an inner product? for $p\neq 2$

I'll give you a hint.

Let $$f(x)=\begin{cases}2,x\in[0,3/4],\\0,x>3/4,\end{cases}\quad g(x)=f(1-x).$$

Then the parallelogram law says that in hilbert spaces we have $$2\|f\|^2+2\|g\|^2=\|f-g\|^2+\|f+g\|^2.$$

Can you calculate the norms above?

How to find these...

By linearity of the integral one has: $$\mathrm{supp}f\cap\mathrm{supp}g=\varnothing:\quad\int|f\pm g|^p\mathrm{d}\lambda=\int|f|^p\mathrm{d}\lambda+\int|g|^p\mathrm{d}\lambda$$

So take a block and shift it slightly: $$f:=\chi_{[0,1]}:\quad f_\varepsilon(x):=f(x-\varepsilon)$$

Then for an appropriate choice: $$\|f+f_\varepsilon\|_p^2+\|f-f_\varepsilon\|_p^2\neq2\|f\|_p^2+2\|f_\varepsilon\|_p^2$$

...besides, this idea should work for any nontrivial function not only blocks.

For the case in which $1 \leq p < \infty$:

Let $\displaystyle f = \bigg( \frac{1}{\mu(A)} \bigg)^{1/p} \cdot \chi_A$ and $\displaystyle g = \bigg( \frac{1}{\mu(B)} \bigg)^{1/p} \cdot \chi_B$ where $A, B$ both have nonzero finite measure, are disjoint and $\chi$ is the indicator function.

Notice that $$\|f\|_p = \bigg( \int \vert f \vert^p \, d\mu \bigg)^{1/p} = \bigg( \int \frac{1}{\mu(A)} \chi_A \, d\mu \bigg)^{1/p} = \bigg( \int_A \frac{1}{\mu(A)} \, d\mu \bigg)^{1/p} = 1$$ and similarly $\|g\|_p = 1$.

Observe that

$$\begin{align*} \|f + g\|_p&= \bigg( \int \vert f + g \vert^p \, d\mu \bigg)^{1/p} \\ &=\bigg( \int \bigg \vert \bigg( \frac{1}{\mu(A)} \bigg)^{1/p} \cdot \chi_A + \bigg( \frac{1}{\mu(B)} \bigg)^{1/p} \cdot \chi_B \bigg \vert^p \, d\mu \bigg)^{1/p} \\ &= \bigg( \int \frac{1}{\mu(A)} \chi_A + \frac{1}{\mu(B)} \chi_B \, d\mu \bigg)^{1/p} \\ &= \bigg( \int_A \frac{1}{\mu(A)} \, d\mu + \int_B \frac{1}{\mu(B)} \, d\mu \bigg)^{1/p} \\ &= 2^{1/p} \end{align*}$$

Note: We use the fact that $A$ and $B$ are disjoint and get $\chi_A \cdot \chi_B = 0$ when moving from line 2 to line 3 above.

Similar reasoning shows $\|f - g\|_p = 2^{1/p}$.

Recall that in an inner product space, the Parallelogram law holds: $\|f + g\|_p^2 + \|f - g\|_p^2 = 2 \|f\|_p^2 + 2 \|g\|_p^2$.

But $\|f + g\|_p^2 + \|f - g\|_p^2 = 2 \cdot 2^{2/p}$ and $2 \|f\|_p^2 + 2 \|g\|_p^2 = 4$ which means that we must have $2^{2/p} = 2$ which only happens if $p = 2$.

For the case in which $p = \infty$:

Let $f = \chi_A$ and $g = \chi_B$ where $A, B$ are disjoint and both have nonzero measure.

Notice that $\|f\|_\infty = \|g\|_\infty = 1$ and $\|f + g\|_\infty = \|f - g\|_\infty = 1$. The Parallelogram Law doesn't hold because $\|f + g\|_\infty^2 + \|f - g\|_\infty^2 = 2$ but $ 2 \|f\|_\infty^2 + 2 \|g\|_\infty^2 = 4$