I think it is not a metric for the space because if you take $f(x)=x$ a polynomial in $P([0,1])$ then I can show that:
$d(x,x) = \int_{0}^{1} x^{2}dx=\frac{x^{3}}{3} \big|_{0}^{1}=\frac{1}{3} \neq 0$
I'm doubting about this because the text says that the inner product given by $\int_{0}^{1}f(x)g(x)$ induces a metric for P([0,1]).
You're right, for the right reason, that $d(f, g)$ is not a metric. However, your text is also right that $\langle f, g\rangle = \int_0^1 f(x) g(x) \, \mathrm{d}x$ forms an inner product on $P[0, 1]$. This is not the same as a distance function; it's analogous to the dot product on $\Bbb{R}^n$. We can use it to define a norm: $$\|f\| = \sqrt{\langle f, f \rangle} = \left(\int_0^1 f(x)^2 \, \mathrm{d}x\right)^{\frac{1}{2}}.$$ Then, from this norm, we can define a metric in the usual way: $$d(f, g) = \|f - g\| = \left(\int_0^1 (f(x) - g(x))^2 \, \mathrm{d}x\right)^{\frac{1}{2}}.$$ So, no, not a metric, but yes, it does induce a (different) metric, as an inner product.