[proofplan] Fix a point $x \in X$ and regard its forward orbit as a sequence in the compact [metric space](/page/Metric%20Space) $X$. Compactness of a metric space implies [sequential compactness](/page/Sequential%20Compactness), so this sequence has a convergent subsequence. The limit of that subsequence lies in $X$ and, by the subsequential characterization of accumulation points of sequences, is an accumulation point of the forward orbit. [/proofplan] [step:Form the forward orbit sequence in $X$] Fix $x \in X$. Define the sequence $a: \{0,1,2,\dots\} \to X$ by \begin{align*} a_n := f^n(x) \end{align*} for every integer $n \ge 0$. Since $f: X \to X$ and $x \in X$, induction on $n$ gives $a_n \in X$ for every integer $n \ge 0$. Thus $(a_n)_{n \ge 0}$ is a sequence in the compact metric space $(X,d)$. [/step] [step:Extract a convergent subsequence by sequential compactness] Because $(X,d)$ is a compact metric space, every sequence in $X$ has a convergent subsequence. Applying this to $(a_n)_{n \ge 0}$, there exist a point $y \in X$ and a strictly increasing sequence $(n_k)_{k \in \mathbb N}$ of nonnegative integers such that \begin{align*} \lim_{k \to \infty} d(a_{n_k}, y) = 0. \end{align*} Since $a_{n_k} = f^{n_k}(x)$ for every $k \in \mathbb N$, this is exactly \begin{align*} \lim_{k \to \infty} d(f^{n_k}(x), y) = 0. \end{align*} [guided] We now use the compactness hypothesis in its sequential form. The sequence under consideration is not an arbitrary object outside the space: from the previous step, each term $a_n = f^n(x)$ lies in $X$. Therefore the sequential [compactness theorem](/theorems/2748) for compact metric spaces applies to $(a_n)_{n \ge 0}$. That theorem gives two pieces of data: first, a strictly increasing sequence $(n_k)_{k \in \mathbb N}$ of nonnegative integers, which selects a subsequence; second, a point $y \in X$ such that the selected terms converge to $y$ in the metric $d$. Written out, this says \begin{align*} \lim_{k \to \infty} d(a_{n_k}, y) = 0. \end{align*} Finally, the definition of the sequence $a_n$ was $a_n := f^n(x)$. Substituting this definition into the convergence statement gives \begin{align*} \lim_{k \to \infty} d(f^{n_k}(x), y) = 0. \end{align*} This is the precise subsequential convergence statement needed to produce an accumulation point of the forward orbit. [/guided] [/step] [step:Identify the subsequential limit as an accumulation point of the orbit] By definition, a point $y \in X$ is an accumulation point of the sequence $(f^n(x))_{n \ge 0}$ if there is a strictly increasing sequence $(n_k)_{k \in \mathbb N}$ of nonnegative integers such that $f^{n_k}(x) \to y$ in $(X,d)$. The subsequence extracted above satisfies precisely this condition. Hence $y$ is an accumulation point of the forward orbit sequence $(f^n(x))_{n \ge 0}$. Since $x \in X$ was arbitrary, every forward orbit sequence in $X$ has an accumulation point. The continuity of $f$ is not used in this argument, but it is compatible with the stated hypotheses. [/step]