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[Editing...see the comments below]

As I stated in the comments, the problem is usually that such topologies do not distinguish between cones and universal cones. However, in the special case where the categories are preorders (i.e., categories where each $Hom(A, B)$ has at most one arrow) arrow), then it is possible to give a topology on the set of objects in such a way that for complete preorders, a functor is continuous with respect to the topologies considered if and only if it preserves limits (i.e., inf) of totally (pre)ordered diagrams. Let us call a functor between complete preorders weakly continuous if and only it does exactly that, i.e., if it is continuous with respect preserves all limits corresponding to the topologies consideredtotally (pre)ordered diagrams. We can topologize these categories the preorders as follows:

Consider first the category $\mathbf{2}$ consisting of an arrow $0 \to 1$ between two objects. Define a topology there which has the object $1$ as the non-trivial closed set. For a general preorder $\mathcal{C}$, define now a topology as the one whose closed sets are of the form $F^{-1}(1)$ for limit-preserving weakly continuous functors $F: \mathcal{C} \to \mathbf{2}$. To see that it is indeed a topology, note that if $(A_i)_i$ are closed sets corresponding to the preimages of $1$ of the limit preserving weakly continuous functors $(F_i)_i$, respectively, then $\prod_i F_i$ is a limit preserving weakly continuous functor such that the preimage of $1$ is exactly $\cap_i A_i$. Similarly, if $A, B$ are closed sets corresponding to the preimages of $1$ of the limit preserving weakly continuous functors $F, G$, respectively, then $F \coprod G$ is limit preserving weakly continuous and the preimage of $1$ is exactly $A \cup B$ (note that in general the coproduct of two limit-preserving limit preserving functors from preorders to $\mathbf{2}$ need not be limit preserving, but it is in the case of functors from preorders so we do need to $\mathbf{2}$).restrict our considerations to weakly continuous functors).

It is clear that if $F: \mathcal{C} \to \mathcal{D}$ preserves limitsis weakly continuous, it is continuous with respect to the topologies defined above, since for a closed set $A$ in $\mathcal{D}$, preimage of $1$ of the functor $G$, we have that $F^{-1}(A)$ is the preimage of $1$ of the limit preserving functor weakly continuous composition $GF$. Conversely, let us see that if $F$ is continuous with respect to the topologies then it must preserve limitsbe weakly continuous. If $(C \to C_i)_i$ is a limiting cone in $\mathcal{C}$, \mathcal{C}$ corresponding to a totally (pre)ordered diagram $(C_i)_i$, then by definition $C$ belongs to the closure of $(C_i)_i$, and hence $F(C)$ must belong to the closure of $(F(C_i))_i$ in $\mathcal{D}$. If $(F(C) \to F(C_i))_i$ were not a universal cone, let $D$ be the vertex of such a cone; we have an induced arrow $F(C) \to D$ (and hence no arrow $D \to F(C)$). But then the representable functor $[D, -]$ (regarded as a functor with values in $\mathbf{2}$) would be weakly continuous (in fact, limit preservingpreserving), and the closed set which is the preimage of $1$ contains each $F(C_i)$ but not $F(C)$, which is absurd, since in that case $F(C)$ would not belong to such a closed set containing the $F(C_i)$. Therefore, $(F(C) \to F(C_i))_i$ must be a universal cone and the proof is complete.
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[Editing...see the comments below]

As I stated in the comments, the problem is usually that such topologies do not distinguish between cones and universal cones. However, in the special case where the categories are preorders (i.e., categories where each $Hom(A, B)$ has at most one arrow) then it is possible to give a topology on the set of objects in such a way that for complete preorders, a functor preserves limits (i.e., inf) if and only if it is continuous with respect to the topologies considered. We can topologize these categories as follows:

Consider first the category $\mathbf{2}$ consisting of an arrow $0 \to 1$ between two objects. Define a topology there which has the object $1$ as the non-trivial closed set. For a general preorder $\mathcal{C}$, define now a topology as the one whose closed sets are of the form $F^{-1}(1)$ for limit-preserving functors $F: \mathcal{C} \to \mathbf{2}$. To see that it is indeed a topology, note that if $(A_i)_i$ are closed sets corresponding to the preimages of $1$ of the limit preserving functors $(F_i)_i$, respectively, then $\prod_i F_i$ is a limit preserving functor such that the preimage of $1$ is exactly $\cap_i A_i$. Similarly, if $A, B$ are closed sets corresponding to the preimages of $1$ of the limit preserving functors $F, G$, respectively, then $F \coprod G$ is limit preserving and the preimage of $1$ is exactly $A \cup B$ (note that in general the coproduct of two limit-preserving functors need not be limit preserving, but it is in the case of functors from preorders to $\mathbf{2}$).

It is clear that if $F: \mathcal{C} \to \mathcal{D}$ preserves limits, it is continuous with respect to the topologies defined above, since for a closed set $A$ in $\mathcal{D}$, preimage of $1$ of the functor $G$, we have that $F^{-1}(A)$ is the preimage of $1$ of the limit preserving functor $GF$. Conversely, let us see that if $F$ is continuous then it must preserve limits. If $(C \to C_i)_i$ is a limiting cone in $\mathcal{C}$, then by definition $C$ belongs to the closure of $(C_i)_i$, and hence $F(C)$ must belong to the closure of $(F(C_i))_i$ in $\mathcal{D}$. If $(F(C) \to F(C_i))_i$ were not a universal cone, let $D$ be the vertex of such a cone; we have an induced arrow $F(C) \to D$ (and hence no arrow $D \to F(C)$). But then the representable functor $[D, -]$ (regarded as a functor with values in $\mathbf{2}$) would be limit preserving, and the closed set which is the preimage of $1$ contains each $F(C_i)$ but not $F(C)$, which is absurd, since in that case $F(C)$ would not belong to such a closed set containing the $F(C_i)$. Therefore, $(F(C) \to F(C_i))_i$ must be a universal cone and the proof is complete.
show/hide this revision's text 2 added 134 characters in body

As I stated in the comments, the problem is usually that such topologies do not distinguish between cones and universal cones. However, in the special case where the categories are preorders (i.e., categories where each $Hom(A, B)$ has at most one arrow) then it is possible to give a topology on the set of objects in such a way that for complete preorders, a functor preserves limits (i.e., inf) if and only if it is continuous with respect to the topologies considered. We can topologize these categories as follows:

Consider first the category $\mathbf{2}$ consisting of an arrow $0 \to 1$ between two objects. Define a topology there which has the object $1$ as the non-trivial closed set. For a general preorder $\mathcal{C}$, define now a topology as the one whose closed sets are of the form $F^{-1}(1)$ for limit-preserving functors $F: \mathcal{C} \to \mathbf{2}$. To see that it is indeed a topology, note that if $A, B$ (A_i)_i$ are closed sets corresponding to the preimages of $1$ of the limit preserving functors $F, G$, (F_i)_i$, respectively, then $F \prod G$ \prod_i F_i$ is a limit preserving functor such that the preimage of $1$ is exactly $A \cap B$\cap_i A_i$. Similarly, if $A, B$ are closed sets corresponding to the preimages of $1$ of the limit preserving functors $F, G$, respectively, then $F \coprod G$ is limit preserving and the preimage of $1$ is exactly $A \cup B$ (note that in general the coproduct of two limit-preserving functors need not be limit preserving, but it is in the case of functors from preorders to $\mathbf{2}$).

It is clear that if $F: \mathcal{C} \to \mathcal{D}$ preserves limits, it is continuous with respect to the topologies defined above, since for a closed set $A$ in $\mathcal{D}$, preimage of $1$ of the functor $G$, we have that $F^{-1}(A)$ is the preimage of $1$ of the limit preserving functor $GF$. Conversely, let us see that if $F$ is continuous then it must preserve limits. If $(C \to C_i)_i$ is a limiting cone in $\mathcal{C}$, then by definition $C$ belongs to the closure of $(C_i)_i$, and hence $F(C)$ must belong to the closure of $(F(C_i))_i$ in $\mathcal{D}$. If $(F(C) \to F(C_i))_i$ were not a universal cone, let $D$ be the vertex of such a cone; we have an induced arrow $F(C) \to D$ (and hence no arrow $D \to F(C)$). But then the representable functor $[D, -]$ (regarded as a functor with values in $\mathbf{2}$) would be limit preserving, and the closed set which is the preimage of $1$ contains each $F(C_i)$ but not $F(C)$, which is absurd, since in that case $F(C)$ would not belong to such a closed set containing the $F(C_i)$. Therefore, $(F(C) \to F(C_i))_i$ must be a universal cone and the proof is complete.
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