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 $\mathbf{Top}$.)

fullsubcategory of $Top$, or can you be more flexible? I can tell you now that monomorphisms in $Top$ given by subspaces are the same thing asregularmonomorphisms, i.e., monomorphisms given as equalizers of some pair of arrows. There are various surrogates of $Top$ which might be worth considering, e.g., the topological topos (which contains sequential spaces as a full subcategory), but to answer the question better, it would help to know how flexible you are with regard to notions of space. – Todd Trimble♦ Jun 22 '11 at 21:08