Theorem: suppose $G$ is a topological group, $E$ a Banach Space, and $\pi$ a representation of $G$ on $E$. Suppose that for all $x\in E$ the map $g\to \pi(g)x$ is continuous: $G\to E$ when $E$ has the norm topology. Finally, suppose that $K^*$ is a weak-$*$ compact subset of $E^*$ that is invariant under $\pi^*.$ Then, there is a functional $\psi\in K^*$ that is fixed under $\pi.$
The proof proceeds as follows: If $\mathcal F$ is the collection of convex, weak-$^*$ closed subsets of $K^*$ invariant under $\pi^*,$ Zorn's Lemma gives us a member of $\mathcal F,$ which we also call $K^*$ for convenience, minimal with this property. The idea then is to show that $K^*$ is a singleton. If not, there is an $x_0\in E,\ \psi_1,\ \psi_2\in K^*$ such that $\psi_1(x_0)\neq \psi_2(x_0).$ Now define $p(\psi)=\sup_{g\in G}|\psi(\pi(g))x_0)|$
$K^*$ is closed, hence compact, and an application of Banach-Steinhaus shows it is norm bounded. Using this fact, one then shows that $p$ is continuous with respect to the weak$^*$ topology so the sets $B(\eta.r)=\left \{ \psi\in K^{*}:p(\psi-\eta)<r \right \}$ are open and convex (since $p$ is positive and subadditive).
Now define $d=\sup\left \{ p(\phi-\psi):\phi,\psi\in K^* \right \}.\ d>0$ because $p(\psi_1-\psi_2)>0).$
There are $\left \{ \psi_k \right \}^{n}_{k=1}$ such that $\left \{ B_k(\psi_k,d/2) \right \}^{n}_{k=1}$ is a finite cover of $K^*$ and the fact that $K^*$ is convex implies that $\psi^{*}=\frac{1}{n}\sum_{k=1}^{n}\psi_k\in K^*.$
Now let $\psi\in K^*.$ Then, $\psi \in B_k(\psi_k,d/2)$ for some $1\le k\le n$ so wlog $k=1$ and then $p(\psi-\psi^*)\le \frac{d}{2}+\frac{1}{n}p((\psi_1-\psi_2)+(\psi_1-\psi_3)+\cdots +(\psi_1-\psi_n))\le\frac{d}{2}+\frac{(n-1)d}{n}=d'$
and it is here that Royden claims that $d'<d$, which is clearly not true.
The rest of the proof, which may be found here in chapter 22, is clear but relies heavily on what appears to be this false claim, which I have not been able to patch up.
This is a typo - it should read $d'=\frac1n((n-1)d+\frac d2) $, which is indeed less than $d$.