= Solution
We first prove the compact-domain <Schauder fixed-point theorem>. Let $K$ be a nonempty compact <convex set> in a <Banach space> $X$, and let $T:K\to K$ be a <continuous map>. The <Brouwer fixed-point theorem> states that every <continuous map> of a finite-dimensional <closed unit ball> into itself has a <fixed point>. By the <homeomorphism> proved in part (a)(ii), it applies equally to any nonempty finite-dimensional compact <convex set>, including a singleton.
For $\varepsilon>0$, use the supplied <Schauder approximation lemma> to obtain a finite-dimensional <linear subspace> $L_\varepsilon$ and a <continuous map> $T_\varepsilon:K\to K\cap L_\varepsilon$ with $\|Tx-T_\varepsilon x\|\leq\varepsilon$. The set $K_\varepsilon=K\cap L_\varepsilon$ is nonempty because it contains $T_\varepsilon(K)$. It is a compact <convex set>, since a finite-dimensional <linear subspace> is closed. Restricting $T_\varepsilon$ to $K_\varepsilon$ and applying the <Brouwer fixed-point theorem> gives $x_\varepsilon=T_\varepsilon x_\varepsilon$, and thus
$$
\|Tx_\varepsilon-x_\varepsilon\|\leq\varepsilon.
$$
Take $\varepsilon_j=1/j$. The <compactness> of $K$ yields a <convergent subsequence> $x_{\varepsilon_{j_m}}\to x_*\in K$. The <continuity> of $T$ and the preceding estimate give
$$
\boxed{Tx_*=x_*.}
$$
For completeness, this also proves the version on a nonempty closed bounded <convex set> $C$ when $T(C)$ is <relatively compact>. Let $K=\overline{\operatorname{conv}}T(C)\subset C$. The <closed convex hull of a compact set> in a <Banach space> is compact. To verify the fact needed here, cover $T(C)$ by finitely many <balls> of radius $\delta$. Replacing every term of a <convex combination> by a centre changes it by at most $\delta$ in <norm>. The <convex hull> of those finitely many centres is compact in a finite-dimensional <linear subspace>, so it has a <finite net> of radius $\delta$. Thus $\operatorname{conv}T(C)$ is <totally bounded>. Its closure is <complete> and <totally bounded>, hence compact. Now $T(K)\subset T(C)\subset K$, so the compact-domain result supplies a <fixed point> in $C$.
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