Possible mistake in the solution of Baby Rudin Ch. 6 Ex. 11 This is exercise 11, Chapter 6 in Baby Rudin:

Let $\alpha$ be a fixed increasing function on $[a, b]$. For $u \in \mathscr{R}(\alpha)$, define
$$ \lVert u \rVert_2 = \left\{ \int_a^b \lvert u \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2}. $$
Suppose $f, g, h \in \mathscr{R}(\alpha)$, and prove the triangle inequality
$$ \lVert f-h \rVert_2 \leq  \lVert f-g \rVert_2 + \lVert g-h \rVert_2 $$
as a consequence of the Schwarz inequality

where by "Schwarz inequality," Rudin means the following version of Holder's inequality:

$$\left\lvert \int_a^b f g \ \mathrm{d} \alpha \right\rvert \leq \left\{ \int_a^b \lvert f \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2} \left\{ \int_a^b \lvert g \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2}. $$

The solution to this question in the solutions manual is:

In the first display on the left-hand side, it should read $||f-h||_2^2$ instead of $||f-h||_2^1$. My question is about the only inequality in the proof. In particular, how is it true that $$\int_a^b |f-g||g-h| \ d\alpha \le \left\{ \int_a^b \lvert f-g \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2} \left\{ \int_a^b \lvert g-h \rvert^2 \ \mathrm{d} \alpha \right\}^{1/2}?$$ This does not looks like the proper usage of the "Schwarz inequality" or am I missing something? Can someone please propose a correct version of the solution?
The solution to the same question here seems to suffer from the same issue, if this is an issue at all, that is.
 A: Regarding your first question I agree, it seems it should be $\|g-h\|_{2}^{2}$ on the LHS.
For your second question, the Cauchy-Schwarz inequality is a special case of H$\ddot{\text{o}}$lder's inequality in the sense that it is the case that is applicable to inner products. So in the second display you have which explains this the intermediate steps are,
\begin{align}
\bigg|\int_{a}^{b}fgd\alpha\bigg|\leq\int_{a}^{b}|f||g|d\alpha=\|fg\|_{1}\leq\|f\|_{2}\|g\|_{2}=\bigg(\int_{a}^{b}|f|^{2}d\alpha\bigg)^{1/2}\bigg(\int_{a}^{b}|g|^{2}d\alpha\bigg)^{1/2}
\end{align}
The other thing to note here is it seems you are misreading the solution. You are asking how,
\begin{align}
\int_{a}^{b}|f-g||g-h|d\alpha\leq\|f\|_{2}\|g\|_{2},
\end{align}
which is not what is being presented in the proof. In Rudin's proof he is using,
\begin{align}
\int_{a}^{b}|f-g||g-h|d\alpha\leq\|f-g\|_{2}\|g-h\|_{2}.
\end{align}
EDIT: To clarify the last inequality, since $f,g,h$ are all Riemann integrable with respect to $d\alpha$ then $f-g$ and $g-h$ are also Riemann integrable. So consider,
\begin{align}
\int_{a}^{b}|f-g||g-h|d\alpha=\|(f-g)(g-h)\|_{1}\leq\|f-g\|_{2}\|g-h\|_{2}=\bigg(\int_{a}^{b}|f-g|^{2}d\alpha\bigg)^{1/2}\bigg(\int_{a}^{b}|g-h|^{2}d\alpha\bigg)^{1/2}
\end{align}
EDIT2: As Paramanand pointed out, since $f-g$ and $g-h$ are Riemann integrable replace $f$ by $|f-g|$ and $g$ by $|g-h|$ in the given Cauchy-Schwarz inequality to get,
\begin{align}
\int_{a}^{b}|f-g||g-h|d\alpha=\bigg|\int_{a}^{b}|f-g||g-h|d\alpha\bigg|\leq\bigg(\int_{a}^{b}|f-g|^{2}d\alpha\bigg)^{1/2}\bigg(\int_{a}^{b}|g-h|^{2}d\alpha\bigg)^{1/2}
\end{align}
