My linear algebra textbook proposes $<f,g> = \int_{0}^{1} {f(x)g(x)}dx$ as an inner product on the space $V = C[0,1]$, the set of continuous functions on the interval $[0,1]$.
The textbook mentions that continuity is used to prove the property that $<f,f>\geq 0$, with equality holding iff $f=0$.
My question is how is continuity used to prove that last part?
You are basically looking to prove that for a continuous function $f\colon[0,1]\to \mathbb{R}$, we have $\int_{0}^{1}f(x)^{2}dx\geq 0$, with equality holding iff $f\equiv 0$. Since $f(x)^{2}\geq 0 $ for all $x$, we will always have $\int_{0}^{1}f(x)^{2}dx\geq 0$ (continuity of $f$ not needed here).
Now, suppose that the integral is $0$. Suppose on contrary that $f(y)\not=0$ for some $y\in[0,1]$. Then, $f(y)^{2}>0$. Since $f^{2}$ is continuous, there is some interval $(a,b)\subset[0,1]$ containing $y$ such that $f(x)^{2}>\frac{1}{2}f(y)^{2}$ for all $x\in (a,b)$. Thus, $$ \begin{aligned} \int_{0}^{1}f(x)^{2}dx&=\int_{0}^{a}f(x)^{2}dx+\int_{a}^{b}f(x)^{2}dx+\int_{b}^{1}f(x)dx\\ &\geq 0 +\int_{a}^{b}\frac{1}{2}f(y)^{2}dx+0\\ &=\frac{1}{2}(b-a)f(y)^{2}\\ &>0, \end{aligned} $$ which is a contradiction.