On a nonempty compact metric domain, the compact-open topology is the uniform topology

A uniform ball of radius $0<\delta <\min \{\epsilon ,1\}$ about $f$ is contained in every compact-convergence basic set $B_K(f,\epsilon )$, because its inequality holds at every point of $X$.

Conversely take $0<\epsilon <1$ and $g\in B_X(f,\epsilon )$, so $d(f(x),g(x))<\epsilon$ at every $x\in X$. Since $X$ is nonempty and compact, clause (U3) of [L2] makes $M:={\max }_{x\in X}d(f(x),g(x))$ exist, and $M<\epsilon <1$ because the maximum is one of the values. Hence $\bar{\rho }(f,g)={\sup }_{x\in X}\min \{d(f(x),g(x)),1\}=M<\epsilon$, so $g$ lies in the uniform ball of radius $\epsilon$. For the reverse inclusion note that step 1.1 is stated for a radius strictly below $\epsilon$ and so cannot be instantiated at $\epsilon$ itself; argue directly instead: if $\bar{\rho }(f,g)<\epsilon <1$ then $\min \{d(f(x),g(x)),1\}<\epsilon <1$ at every $x$, so $d(f(x),g(x))<\epsilon$ at every $x$ and $g\in B_X(f,\epsilon )$. So $B_X(f,\epsilon )$ is exactly that uniform ball.

Steps 1.1 and 2.1 show that compact convergence and uniform convergence induce the same topology. By [L1], that topology is also the compact-open topology.