[proofplan] We use compactness to choose a minimal non-empty closed forward-invariant subset. In such a minimal set, every forward tail of every orbit is dense in the minimal set, so each point is accumulated by its own positive iterates. [/proofplan] [step:Choose a minimal compact invariant set] Let $\mathcal{K}$ be the family of non-empty closed subsets $K\subseteq X$ satisfying \begin{align*} T(K)\subseteq K. \end{align*} The family is non-empty because $X\in\mathcal{K}$. Order $\mathcal{K}$ by inclusion. If $(K_i)_{i\in I}$ is a chain in $\mathcal{K}$, then the compactness of $X$ and the finite intersection property imply that \begin{align*} K:=\bigcap_{i\in I}K_i \end{align*} is non-empty and closed. Also $T(K)\subseteq K$, so $K\in\mathcal{K}$. By Zorn's lemma, $\mathcal{K}$ has a minimal element; call it $M$. [/step] [step:Show that positive orbit tails are dense in the minimal set] Choose any $x\in M$. The set \begin{align*} L:=\overline{\{T^n x:n\geq1\}} \end{align*} is non-empty, closed, and contained in $M$. It is forward-invariant because continuity of $T$ gives \begin{align*} T(L)\subseteq \overline{\{T^{n+1}x:n\geq1\}}\subseteq L. \end{align*} By minimality of $M$, we must have $L=M$. [/step] [step:Extract a recurrent point] Since $x\in M=L$, every neighbourhood $U$ of $x$ intersects $\{T^n x:n\geq1\}$. Hence there is some $n\geq1$ such that $T^n x\in U$. Thus $x$ is topologically recurrent. [/step]