By definition of norm induced from inner product, $||x||^2 = \langle x,x\rangle$.
That is, if $||a+bx|| = 1$ then $\langle a+bx,a+bx\rangle = 1$ ,so writing this down:
$$
\int_{0}^1 (a+bx)(a+bx) dx = 1 \implies \int_0^1 (a^2 + 2abx + b^2x^2) dx = 1 \\ \implies a^2 + ab + \frac {b^2}3 = 1
$$
therefore, the statement that $a^2 + b^2 = 1$ is FALSE.
This is actually quite clear with an example : $a=0 , b=1$ satisfies $a^2+b^2 = 1$, and gives the polynomial $x$, but $||x||^2 = \int_0^1 x^2 = \frac 13$, so $||x|| \neq 1$. Instead, the other statement given is correct. Replacing $a$ and $b$ by $c$ and $d$ gives you the other analogously correct statement.
Once this happens, you may set $a,b$ to any suitable values, and check what happens to $c$ and $d$.
For example, set $b = 0$ : from the above equation, this forces $a = \pm 1$, we will take $a =1$.
From the equation that $\langle a+bx,c+dx\rangle = 0$ that the author has derived in your question above, substituting (and cancelling $b$) and rearranging gives $2c+d = 0$, so $d = -2c$.
This must be combined with $c^2 + cd + \frac{d^2}{3} = 1$. Setting it, we get $c^2(1 - 2 + \frac 43) = 1$, so $c^2 = 3$. Just take $c = + \sqrt 3$, so $d = -2\sqrt 3$.
In this manner, we may verify that the polynomial $a+bx = 1$ and $c + dx = \sqrt 3(1 - 2x)$ form an orthonormal basis for the space.
First we show that $\left<e_n,e_m\right>=0$ for $n\neq m$.
We do this by considering
$$\left<e_n,e_m\right> = \sum_{k=1}^\infty e_n^ke_m^k.$$
Remark that $e_n^k$ is only nonzero if $k=n$, and likewise $e_m^k$ is only nonzero if $k=m$. It follows that in every term, at least one of the factors is zero, since $n\neq m$. By consequence, the sum evaluates to zero.
The following proof for orthonormality illustrates how we apply this.
For $f_{2n}$ and $f_{2m}$ with $n < m$ we get
$$
\begin{align*}
\left<f_{2n},f_{2m}\right> = {} &\frac12\left<e_{2n-1} + e_{2n},e_{2m-1} + e_{2m}\right> \\
= {} & \frac12\left(\left<e_{2n-1},e_{2m-1}\right> + \left<e_{2n},e_{2m-1}\right> + \left<e_{2n-1},e_{2m}\right> + \left<e_{2n},e_{2m}\right>\right).
\end{align*}
$$
Because $n<m$, the operands are orthogonal in every scalar product, which means every scalar product is zero.
I will do the calculations for the third product, since it's slightly less trivial. Since $n<m$ we have $n\leq m-1$, which means $2n\leq 2m-2 < 2m-1$. Therefore, $e_{2n}$ is not equal (and hence is orthogonal) to $e_{2m-1}$. For the other cases, we make the same inequality argument, but finding the needed inequality is easier.
For $f_{2n}$ and $f_{2m-1}$ with $n < m$, we get the same four scalar products, but the last two will be prepended by a minus sign, which doesn't invalidate the above argument.
For $f_{2n-1}$ and $f_{2n}$ with $n < m$ we get
$$
\begin{align*}
\left<f_{2n-1},f_{2n}\right> = {} &\frac12\left<e_{2n-1} - e_{2n},e_{2n-1} + e_{2n}\right> \\
= {} & \frac12\left(\left<e_{2n-1},e_{2n-1}\right> - \left<e_{2n},e_{2n-1}\right> + \left<e_{2n-1},e_{2n}\right> - \left<e_{2n},e_{2n}\right>\right) \\
= {} & \frac12(1+0+0-1) = 0.
\end{align*}
$$
We have now proven that the $f_k$ are orthogonal. To complete the argument, we consider
$$
\begin{align*}
\left<f_{n},f_{n}\right> = {} &\frac12\left<e_{n-1} \pm e_{n},e_{n-1} \pm e_{n}\right> \\
= {} & \frac12\left(\left<e_{n-1},e_{n-1}\right> \pm \left<e_{n-1},e_{n}\right> \pm \left<e_{n},e_{n-1}\right> + \left<e_{n},e_{n}\right>\right) \\
= {} & \frac12(1 \pm 0 \pm 0 + 1) = 1.
\end{align*}
$$
Best Answer
you have three vectors $$(0,4/5,3/5)^T, (0, 3/5, -4/5)^T, (1, 0, 0)^T $$ with the usual inner product. each of them have length $1$ and they are mutually orthogonal, therefore they form an orthonormal basis for $P_2.$