For a metric domain and a metric target the compact-open topology on $C(X,Y)$ is the topology of compact convergence

First half: let $K \subseteq X$ be compact, let $V \subseteq Y$ be open and let $f \in S(K,V)$; it suffices to produce a real $\varepsilon > 0$ with $B_K(f,\varepsilon) \subseteq S(K,V)$, since then every point of $S(K,V)$ lies in a basic set of $\mathcal{T}_{\mathrm{cc}}$ inside it.

Second half: let $K \subseteq X$ be compact, let $f_0 \in C(X,Y)$, let $\varepsilon_0 > 0$ be real and let $g \in B_K(f_0,\varepsilon_0)$; it suffices to produce a finite intersection of sets $S(K',V')$ containing $g$ and contained in $B_K(f_0,\varepsilon_0)$.

In step 1.1, if $K = \varnothing$ or $V = Y$ then $S(K,V) = C(X,Y)$, which is open in $\mathcal{T}_{\mathrm{cc}}$, and $\varepsilon := 1$ serves since $B_K(f,1) \subseteq C(X,Y)$; so assume $K \ne \varnothing$ and $V \ne Y$, whence $Y \setminus V$ is nonempty.

In step 1.2, if $K = \varnothing$ then $B_K(f_0,\varepsilon_0) = C(X,Y) = S(\varnothing, Y)$, which is subbasic and hence open in $\mathcal{T}_{\mathrm{co}}$; so assume $K \ne \varnothing$.

Under step 2.1: $f[K]$ is a nonempty compact subset of $Y$, and the function $\psi(y) := d(y, Y \setminus V)$ is defined and continuous on $Y$.

Under step 2.2: $M := \max_{x \in K} d(f_0(x), g(x))$ exists and satisfies $M < \varepsilon_0$, because $g \in B_K(f_0,\varepsilon_0)$ makes $\varepsilon_0$ a strict upper bound of the values; put $\delta := \varepsilon_0 - M > 0$.

Under step 2.1: the restriction of $\psi$ to the nonempty compact metric subspace $f[K]$ is continuous, so it attains a least value $\varepsilon := \psi(y_0)$ at some $y_0 \in f[K]$, and $\varepsilon \le \psi(y)$ for every $y \in f[K]$.

Under step 2.2: let $\mathcal{P}$ be the set of pairs $(a,r)$ with $a \in K$, $r > 0$ real and $g[\bar B(a,r) \cap K] \subseteq B(g(a), \delta/4)$; the family $(B(a,r))_{(a,r) \in \mathcal{P}}$ consists of open subsets of $X$ and covers $K$, since continuity of $g$ at $a \in K$ gives $s > 0$ with $g[B(a,s)] \subseteq B(g(a),\delta/4)$ and then $r := s/2$ satisfies $\bar B(a,r) \subseteq B(a,s)$, so $(a,r) \in \mathcal{P}$ and $a \in B(a,r)$.

Under step 2.1: $\varepsilon > 0$, because $y_0 \in f[K] \subseteq V$ with $V$ open gives a real $s > 0$ with $B(y_0,s) \subseteq V$, so every $z \in Y \setminus V$ satisfies $d(y_0,z) \ge s$, making $s$ a lower bound of the distances from $y_0$ to $Y \setminus V$ and hence $\psi(y_0) \ge s > 0$.

Under step 2.2: since $K \ne \varnothing$ is compact, there are $n \in \mathbb{N}$ and pairs $(a_0,r_0), \dots, (a_n,r_n) \in \mathcal{P}$ with $K \subseteq B(a_0,r_0) \cup \dots \cup B(a_n,r_n)$; each index is a pair, so the centres and radii come back with the indices and nothing is selected.

Under step 2.1: for $u \in B_K(f,\varepsilon)$ and $x \in K$ we have $d(f(x),u(x)) < \varepsilon \le \psi(f(x))$ by step 4.1, since $f(x) \in f[K]$; were $u(x) \in Y \setminus V$, the distance $d(f(x),u(x))$ would be one of the distances from $f(x)$ to $Y \setminus V$ and hence at least $\psi(f(x))$, which it is not; so $u(x) \in V$.

Under step 2.2: for $j \le n$ put $K_j := \bar B(a_j,r_j) \cap K$ and $V_j := B(g(a_j), \delta/2)$; each $K_j$ is closed in the compact metric space $(K, d_K)$, being the trace on $K$ of the closed set $\bar B(a_j,r_j)$, hence is compact, and each $V_j$ is open in $Y$.

Under step 2.1: step 6.1 holds for every $x \in K$, so $u[K] \subseteq V$ and $u \in S(K,V)$; hence $B_K(f,\varepsilon) \subseteq S(K,V)$, which is what step 1.1 required, and every $S(K,V)$ is open in $\mathcal{T}_{\mathrm{cc}}$, so $\mathcal{T}_{\mathrm{co}} \subseteq \mathcal{T}_{\mathrm{cc}}$.

Under step 2.2: $g \in S(K_j,V_j)$ for every $j \le n$, since $K_j \subseteq \bar B(a_j,r_j) \cap K$ and $(a_j,r_j) \in \mathcal{P}$ give $g[K_j] \subseteq B(g(a_j),\delta/4) \subseteq V_j$; so $g \in O := S(K_0,V_0) \cap \dots \cap S(K_n,V_n)$, a finite intersection of subbasic sets and hence open in $\mathcal{T}_{\mathrm{co}}$.

Under step 2.2: let $h \in O$ and $x \in K$; step 5.2 gives $j \le n$ with $x \in B(a_j,r_j)$, so $x \in K_j$, whence $d(h(x),g(a_j)) < \delta/2$ and $d(g(x),g(a_j)) < \delta/4$, so $d(h(x),g(x)) < \delta/2 + \delta/4 < \delta$ by the triangle inequality.

Under step 2.2: therefore $d(f_0(x),h(x)) \le d(f_0(x),g(x)) + d(g(x),h(x)) < M + \delta = \varepsilon_0$ for every $x \in K$, that is $h \in B_K(f_0,\varepsilon_0)$; so $O \subseteq B_K(f_0,\varepsilon_0)$, which is what step 1.2 required.

By step 9.1 every point of every basic set of $\mathcal{T}_{\mathrm{cc}}$ is interior to it in $\mathcal{T}_{\mathrm{co}}$, so every basic set of $\mathcal{T}_{\mathrm{cc}}$ is open in $\mathcal{T}_{\mathrm{co}}$, and since those basic sets generate, $\mathcal{T}_{\mathrm{cc}} \subseteq \mathcal{T}_{\mathrm{co}}$.

With step 7.1 the two inclusions give $\mathcal{T}_{\mathrm{co}} = \mathcal{T}_{\mathrm{cc}}$.