A subcategory of top where subspaces and subobjects coincide?

A supplement to Todd's answer with the observation that the requested property fails in all the usual cartesian closed full subcategories of $\mathbf{Top}$. Indeed, there are even regular subobjects (see Todd's comment for the definition) that are not subspaces. (Todd's own counterexample in compactly generated spaces describes a non-regular subobject that is not a subspace.)

By the "usual" subcategories, I mean those that fall under the general treatment of cartesian closed full subcategories of $\mathbf{Top}$ in:

M. Escardó, J.Lawson, A. Simpson. Comparing Cartesian closed categories of (core) compactly generated spaces. Topology and its Applications, 143 (2004) 105–145.

This considers categories of $\mathcal{C}$-generated spaces for suitable collections $\mathcal{C}$ of topological spaces. (Compactly generated spaces arise by taking $\mathcal{C}$ to be the collection of compact Hausdorff spaces.)

Every such category of $\mathcal{C}$-generated spaces includes the integers $\mathbb{Z}$ with discrete topology. Using cartesian closedness, define the iterated function space $F_2 := \mathbb{Z}^{\mathbb{Z}^\mathbb{Z}}$. I observe below that $F_2$ has a (countable) subspace $Y$ that is not $\mathcal{C}$-generated. Then $Y$ endowed with its $\mathcal{C}$-generated topology gives the promised regular subobject of $F_2$ in the category of $\mathcal{C}$-generated spaces that is not a subspace.

For the observation, first, by Corollary 7.3 of op. cit., the topology of $F_2$ is independent of the choice of $\mathcal{C}$. For convenience, we consider $F_2$ with its topology as a sequential space. It is known that $F_2$ is not a Fréchet–Urysohn space. That is, there is a countable subset $X \subseteq F_2$ and element $x_\infty \in F_2$ such that $x_\infty$ is in the closure of $X$ but is not the limit of any sequence in $X$. (This known fact is fairly easily verified once one has a concrete grasp of $F_2$ as the sequential space of continuous functions from Baire space $\mathbb{Z}^\omega$ to $\mathbb{Z}$.) Define $Y := X \cup \{x_\infty\}$. Then the subspace $Y$ of $F_2$ is (obviously) not sequential, but (by Lemma 6.3 of op. cit.) has a countable pseudobase in (all) the sense(s) of op. cit. Thus, by Theorem 6.10 of op. cit., $Y$ is not $\mathcal{C}$-generated.

Edit. I should have said, Todd's counterexample to the original question also applies to $\mathcal{C}$-generated spaces. The main point of my answer is to show the perhaps more surprising fact that "convenience" (i.e. cartesian closedness) is not even compatible with regular subobjects being subspaces. (This does not conflict with the characterisation of subspaces as regular subobjects in $\mathbf{Top}$, stated by Todd, because equalizers in the category of $\mathcal{C}$-generated spaces are computed differently from in $\mathbf{Top}$.)

Well, I'd say that compact Hausdorff spaces are at least a nice category of spaces where subspaces are equivalent to subobjects. One reason it is nice is that it is a category of algebras for a monad on $Set$, with all the nice properties that entails, for example Barr exactness. It is of course not "convenient" in the technical sense of Steenrod and (earlier) Ronnie Brown: it isn't cartesian closed.

Whatever the candidate is, you absolutely have to worry about whether the subcategory allows the same set to have comparable topologies living as spaces in the same subcategory. For example, in the classically convenient category of compactly generated spaces, you can have for example the real numbers with the standard topology as one object, and the real numbers with the discrete topology as another, and a continuous function $id: \mathbb{R}_{disc} \to \mathbb{R}$. This is certainly a monomomorphism, but certainly not a subspace. So this convenient category doesn't work. (On the other hand, there cannot be two comparable compact Hausdorff topologies on the same set.)

Off-hand I'm having trouble thinking of an example of a full subcategory of $Top$ which is both cartesian closed (and complete and cocomplete) and has this property you are looking for, but perhaps I haven't thought hard enough.