[proofplan] The strategy is to reduce to the affine setting and use transcendence-degree arithmetic. Cover $C$ by affine open subsets; on each, a proper closed subvariety has coordinate ring obtained as a quotient of the coordinate ring of the affine piece. The function field of an irreducible affine curve over $k$ has transcendence degree $1$ over $k$. We show that any proper irreducible closed subvariety has coordinate ring algebraic over $k$, hence (by the Nullstellensatz over algebraically closed $k$) equal to $k$, hence a single point. A finite cover of $C$ by affines plus the irreducible decomposition of $W$ then bound the total number of points by a finite sum. [/proofplan]

[step:Reduce to the case where $W$ is irreducible] A closed subvariety $W \subsetneq C$ — taking "subvariety" to mean a closed subset, possibly reducible — admits a decomposition into irreducible components $W = W_1 \cup \cdots \cup W_r$ (by Noetherian decomposition: $C$ is Noetherian as a variety over $k$, and any descending chain of closed subsets stabilises, giving the finite decomposition). It suffices to show each irreducible $W_i$ is a single point: then $W = \{x_1, \ldots, x_r\}$ is finite of cardinality at most $r$. So assume $W$ is irreducible and $W \subsetneq C$. We show $W$ is a single point. [/step]

[step:Reduce to an affine open subvariety of $C$ containing $W$]Cover $C$ by finitely many affine open subsets: by definition of a variety, $C$ has an open cover by affine varieties, and since $C$ is irreducible (hence Noetherian) we may take a finite such cover $C = U_1 \cup \cdots \cup U_n$ with each $U_i \subset C$ open and isomorphic to an affine variety. Now $W \subseteq C = \bigcup U_i$, and since $W$ is irreducible, $W$ is contained in the closure of $W \cap U_i$ for some $i$ (specifically, the $U_i$ for which $W \cap U_i$ is dense in $W$ — this exists because $W$ is irreducible and the $U_i$ form an open cover). Replace $W$ by $W \cap U_i$ and $C$ by $U_i$: it suffices to show $W \cap U_i$ is finite for each $i$, since then $W$ (being the closure of a finite set) is also finite. Concretely, if we show every irreducible proper closed subvariety of an irreducible affine curve is a point, we are done. Henceforth, assume $C$ is an irreducible affine curve, with coordinate ring $\mathcal{O}_C := k[C]$ — a finitely generated $k$-algebra that is an integral domain (because $C$ is irreducible) of Krull dimension $1$ (because $C$ has dimension $1$).[/step]

[guided]Why does irreducibility of $W$ imply $W \subset \overline{W \cap U_i}$ for some $i$? The open sets $W \cap U_i$ form an open cover of $W$ (in the subspace topology). At least one $W \cap U_i$ is non-empty, since $W \subseteq \bigcup U_i$ and $W$ is non-empty (a single non-empty closed set, and we're handling the empty case separately). For an irreducible space, every non-empty open subset is dense — this is a defining property of irreducibility. So $\overline{W \cap U_i} = W$ for any $i$ with $W \cap U_i \neq \varnothing$. Why does the affine reduction suffice? If we know $W \cap U_i$ is a finite set $\{x_1, \ldots, x_r\}$ for each $i$, then $W$ is a finite union of finite sets — namely $W = \bigcup_i (W \cap U_i)$ — and is itself finite. Why does the dimension descend correctly? An irreducible affine open subvariety of an irreducible curve is again an irreducible affine variety of dimension $1$ (open subsets have the same dimension as the whole variety). So $\mathcal{O}_C$ for this affine $C$ is an integral domain of Krull dimension $1$.[/guided]

[step:Translate to the algebraic setting via the closed-immersion ring map] Let $W \subsetneq C$ be an irreducible proper closed subvariety of the irreducible affine curve $C$. The closed immersion \begin{align*} \iota: W &\hookrightarrow C \end{align*} induces a surjective $k$-algebra homomorphism on coordinate rings: \begin{align*} \iota^*: \mathcal{O}_C &\twoheadrightarrow \mathcal{O}_W \\ f &\mapsto f|_W. \end{align*} The map $\iota^*$ is surjective because $W$ is closed in $C$: by the [algebraic-geometric duality for affine varieties](/theorems/2127), closed immersions of affine varieties correspond to surjective $k$-algebra maps on coordinate rings. Concretely, the kernel of $\iota^*$ is the ideal $I(W) \subset \mathcal{O}_C$ of regular functions on $C$ vanishing on $W$, and $\mathcal{O}_W = \mathcal{O}_C / I(W)$. Since $W$ is irreducible, $I(W) \subset \mathcal{O}_C$ is a prime ideal (by [Irreducibility via Prime Ideals](/theorems/2126), or rather its affine version: a closed subset of an affine variety is irreducible iff its ideal is prime). Since $W \subsetneq C$ is proper, $I(W) \neq (0)$ (else $W = V(0) = C$): there exists a non-zero $g \in I(W)$, i.e.\ $g \in \mathcal{O}_C \setminus \{0\}$ with $\iota^*(g) = 0$. [/step]

[step:Use transcendence-degree arithmetic to show $\mathcal{O}_W$ is algebraic over $k$]The coordinate ring $\mathcal{O}_C$ is an integral domain of Krull dimension $1$. Its field of fractions $K_C := \operatorname{Frac}(\mathcal{O}_C)$, the function field of $C$, has transcendence degree $1$ over $k$ — this is the field-theoretic statement of "$C$ has dimension $1$". We claim $\mathcal{O}_W$ is algebraic over $k$. Suppose for contradiction that $\mathcal{O}_W$ contains an element $t \in \mathcal{O}_W$ transcendental over $k$ — that is, $t$ does not satisfy any non-zero polynomial relation over $k$. Choose a lift $x \in \mathcal{O}_C$ with $\iota^*(x) = t$ (possible because $\iota^*$ is surjective). Then $x$ is also transcendental over $k$, since any non-zero polynomial $P \in k[T]$ with $P(x) = 0$ in $\mathcal{O}_C$ would map to $P(t) = 0$ in $\mathcal{O}_W$, contradicting transcendence of $t$. Now consider the two elements $x, g \in \mathcal{O}_C$ where $g \neq 0$ is the element from Step 3 with $\iota^*(g) = 0$. We claim $x$ and $g$ are algebraically independent over $k$ inside $\mathcal{O}_C$. Suppose for contradiction there is a non-zero polynomial $P \in k[T_1, T_2]$ with $P(x, g) = 0$ in $\mathcal{O}_C$. Write $P = \sum_{r \geq 0} P_r(T_1) T_2^r$ for some polynomials $P_r \in k[T_1]$, where $\deg_{T_2} P = d$ and $P_d \neq 0$. Apply $\iota^*$: \begin{align*} 0 = \iota^*(P(x, g)) = P(\iota^*(x), \iota^*(g)) = P(t, 0) = P_0(t), \end{align*} since $\iota^*(g) = 0$ kills all terms with $T_2^r$ for $r \geq 1$. So $P_0(t) = 0$ in $\mathcal{O}_W$. Since $t$ is transcendental over $k$, this forces $P_0 = 0$ as a polynomial in $k[T_1]$. Consequently $P(T_1, T_2) = T_2 \cdot Q(T_1, T_2)$ for some $Q \in k[T_1, T_2]$ of total degree $\deg P - 1$. Dividing the identity $P(x, g) = 0$ by $g$ in the integral domain $\mathcal{O}_C$ — valid because $g \neq 0$, and integral domains have cancellation — gives $Q(x, g) = 0$. Iterate: by induction on $\deg_{T_2} P$, after $d$ divisions we obtain \begin{align*} P(T_1, T_2) = T_2^d \cdot R(T_1) \end{align*} for some $R \in k[T_1]$, with $R \neq 0$ (since the leading coefficient $P_d \neq 0$ becomes $R$ after fully extracting the $T_2^d$ factor, modulo a non-zero scalar — more precisely, the iteration extracts factors of $T_2$ until the remainder has $T_2$-degree $0$, and the remainder is non-zero because $P$ was non-zero). Substituting $T_1 = x$, $T_2 = g$: \begin{align*} 0 = P(x, g) = g^d R(x). \end{align*} In the integral domain $\mathcal{O}_C$, $g^d \neq 0$ (since $g \neq 0$), so $R(x) = 0$. But $R \in k[T_1]$ is a non-zero polynomial and $x$ is transcendental over $k$, contradiction. Hence $x$ and $g$ are algebraically independent over $k$ in $\mathcal{O}_C$. Their images in $K_C$ remain algebraically independent (the natural map $\mathcal{O}_C \hookrightarrow K_C$ is injective). So the transcendence degree of $K_C / k$ is at least $2$, contradicting $\operatorname{trdeg}_k K_C = 1$. This contradiction shows that no such transcendental $t$ exists: every element of $\mathcal{O}_W$ is algebraic over $k$. So $\mathcal{O}_W$ is integral over $k$.[/step]

[guided]The crux of this step is the inequality $\operatorname{trdeg}_k K_C = 1$, combined with the fact that a non-isomorphic surjection $\iota^*: \mathcal{O}_C \to \mathcal{O}_W$ "uses up" at least one degree of transcendence: it kills a non-zero element $g$, and that element together with any putative transcendental in $\mathcal{O}_W$ would form an algebraically independent pair in $\mathcal{O}_C$. \textbf{Why is $\operatorname{trdeg}_k K_C = 1$?} This is the field-theoretic translation of "$C$ has dimension $1$". For an irreducible affine variety $C \subset \mathbb{A}^N_k$, the dimension equals the Krull dimension of $\mathcal{O}_C$, which equals the transcendence degree of $K_C$ over $k$ (a standard theorem, sometimes called the dimension theorem for affine varieties; it follows from Noether normalisation and the fact that $\mathcal{O}_C$ is a finitely generated $k$-algebra). For dimension $1$, this transcendence degree is exactly $1$. \textbf{Why does the algebraic independence of $x, g$ contradict $\operatorname{trdeg} = 1$?} Algebraically independent elements in $\mathcal{O}_C$ remain algebraically independent in any field extension $K \supseteq \mathcal{O}_C$ — the transcendence degree of a field extension is computed via algebraic independence in any generating set. So $\operatorname{trdeg}_k K_C \geq 2$, contradicting the equality $\operatorname{trdeg}_k K_C = 1$. \textbf{Why does $\iota^*$ kill at least one non-zero element?} Because $\iota^*$ is not injective. Surjective ring maps that are not injective have non-zero kernel — and "non-injective" follows from $W \subsetneq C$ being a proper closed inclusion: if $\iota^*$ were injective, the contravariant equivalence (algebraic-geometric duality) would force $\iota$ to be a dominant morphism, which combined with $W$ being closed and irreducible would force $W = C$. \textbf{Where does the algebraic-closure of $k$ enter?} It is used in Step 5 below, when we conclude from "$\mathcal{O}_W$ is algebraic over $k$" that "$\mathcal{O}_W = k$". Without algebraic closure, $\mathcal{O}_W$ could be a proper finite field extension of $k$.[/guided]

[step:Conclude $\mathcal{O}_W = k$ using algebraic closure] By Step 4, $\mathcal{O}_W$ is integral over $k$ (every element is algebraic over $k$). We claim $\mathcal{O}_W = k$. Since $\mathcal{O}_C$ is a finitely generated $k$-algebra and $\mathcal{O}_W$ is a quotient of $\mathcal{O}_C$, $\mathcal{O}_W$ is also a finitely generated $k$-algebra. A finitely generated $k$-algebra that is an integral domain (which $\mathcal{O}_W$ is, since $W$ is irreducible) and integral over $k$ is a field, finite-dimensional over $k$ (this is a consequence of the going-up theorem: integral extensions preserve "field-ness" of the ground ring; alternatively, integral plus integral domain plus finitely generated implies a finite field extension). Concretely, $\mathcal{O}_W$ is a finite field extension of $k$. Since $k$ is algebraically closed, the only finite (algebraic) field extension of $k$ is $k$ itself. Hence $\mathcal{O}_W = k$. [/step]

[step:Identify $W$ with a single point and conclude] We now show that $\mathcal{O}_W = k$ implies $W$ is a single point. By the [algebraic-geometric duality for affine varieties](/theorems/2127), affine varieties over $k$ correspond to finitely generated $k$-algebras that are integral domains, with morphisms corresponding to $k$-algebra homomorphisms in the opposite direction. The affine variety with coordinate ring $k$ is the single-point variety $\mathbb{A}^0_k = \{*\}$. Hence $W$, having coordinate ring $\mathcal{O}_W = k$, is isomorphic to a single point as a variety. As a subset of $C$, $W$ is therefore a singleton $\{x\}$ for some $x \in C$. Combined with Step 1, the original (possibly reducible) proper closed subvariety $W \subsetneq C$ is a finite union of singletons, hence a finite set of points. Combined with Step 2, the same conclusion holds for any irreducible curve $C$ (not necessarily affine), since $W$ is then a finite union of points coming from finitely many affine pieces $U_i$. This completes the proof. [/step]