Technical issue in the approach to Lie groups taken in a book

Thanks for the question. It is true that I never proved that the simple summands in Theorem 7.8 are themselves semisimple in the sense that I define "semisimple." I would not characterize this omission as a problem, however, since I don't claim they are semisimple. Furthermore, I do not prove the classification of semisimple algebras. (And in any case, the classification would go through the classification of root systems, which decompose in terms of irreducibles.)

Nevertheless, it is an interesting question, and I believe we can prove that each $\mathfrak{g_j}$ is semisimple, using only the methods in Chapter 7 of my book. There are actually two issues, which I am not sure have been clearly separated in the discussion so far. First, we need to show that the decomposition of $\mathfrak{g}$ in Theorem 7.8 actually comes from a decomposition of $\mathfrak{k}$. To establish this, we first observe that if $\mathfrak{g}$ is center-free, then $\mathfrak{k}$ must also be center-free. Thus, $\mathfrak{k}$ will decompose as a direct sum of real, simple algebras $\mathfrak{k_j}$, by the same argument as the proof of Theorem 7.8. But since $\mathfrak{k_j}$ admits an Ad-invariant inner product, the proof of Theorem 7.32 shows that the complexification $\mathfrak{g_j}$ of $\mathfrak{k_j}$ is also simple as a complex Lie algebra. Then by the uniqueness result in Proposition 7.9, these $\mathfrak{g_j}$'s must be actually be the ones in Theorem 7.8.

Second, we need to show that each $\mathfrak{k_j}$ is the Lie algebra of a compact group, which would show that $\mathfrak{g_j}$ is semisimple in the sense used in the book. For this, the argument of suggested by Noah Snyder seems best: Since $\mathfrak{k_j}$ is an ideal, it is invariant under the adjoint action of $K$.The image $K_j$ of $K$ inside $GL(\mathfrak{k_j})$ is compact since it's the continuous image of the compact group $K$. Since $\mathfrak{k_j}$ is center-free, the associated Lie algebra homomorphism (the map sending $X$ to the restriction of $ad_X$ to $\mathfrak{k_j}$) will be zero on each $\mathfrak{k_l}$ with $l\neq j$ but will be injective on $\mathfrak{k_j}$. Thus, the image of the associated Lie algebra map will be isomorphic to $\mathfrak{k_j}$.

Here's a much easier argument more in the spirit of Hall's book.

If K is a compact matrix Lie group with discrete center and its Lie algebra decomposes as $\mathfrak{k}=\mathfrak{a}\oplus \mathfrak{b}$ then look at the image of K under its adjoint action on $\mathfrak{a}$. This is a continuous image of a compact set and so is compact (and closed!) so it gives a compact matrix group A. Since the derivative of Ad is ad, the Lie algebra of A is $\mathfrak{a}$.

As suggested by Vít Tuček, we can take the following approach. Since $K$ is compact and has no center, the Killing form on $\mathfrak{k}$ is negative definite. Thus the restriction of the Killing form to each of the $\mathfrak{k}_i$ is negative definite. We now want to exhibit compact matrix groups $K_i$ with Lie algebra $\mathfrak{k}_i$ using the negative definiteness of the Killing form.

There are many arguments in the literature that negative definiteness implies the existence of a compact group (though about half the ones I found seemed to have a gap). Here's one argument (which I got from Hsiang) which doesn't leave the world of matrix groups.

Suppose $\mathfrak{g}$ is a Lie algebra with negative definite Killing form, and consider $\mathfrak{D}$ its Lie algebra of derivations. A short calculation shows that the Killing form on $\mathfrak{D}$ restricts to the Killing form on $\mathfrak{g}$ and by considering the perpendicular compliment of $\mathfrak{g}$ in $\mathfrak{D}$ it follows that $\mathfrak{g} = \mathfrak{D}$. Hence, $\mathfrak{g}$ is the Lie algebra of the matrix Lie group $\mathrm{Aut}(\mathfrak{g}) \subset \mathrm{GL}(\mathfrak{g})$ (this is clearly a closed subgroup because it is cut out by polynomial equations). This subgroup is compact because it preserves the negative definite Killing form.