Philippe's argument works well. I just wanted to remark that in this special case one does not need to invoke the homotopy exchange property explicitely as long as one already accepts that a model structure with the given cofibrations and fibrant object exists for the category $\mathrm{Top}_\Delta$ of $\Delta$-generated spaces.
Given a map $f\colon X\to Y$ in $\mathrm{Top}_\Delta$, first build the following diagram in $\mathrm{Top}$:
$$ \matrix{ X & \mathop{\longrightarrow}\limits^{\tilde{f}} & N_f & \mathop{\longrightarrow}\limits^{p'} & X \cr {\scriptstyle f} \big\downarrow {\ }& & {\ } \big\downarrow {\scriptstyle f'} & & {\ } \big\downarrow {\scriptstyle f} \cr Y & \mathop{\longrightarrow}\limits_t & Y^I & \mathop{\longrightarrow}\limits_{p_0} & Y \cr & & {\ } \big\downarrow {\scriptstyle p_1} & & \cr & & Y{\ } & & } $$
Here $(Y^I,p_0,p_1,t)$ is the usual path object on $Y$, $N_f$ is the pullback of $p_0$ and $f$, and $\tilde{f}$ is the map induced by $id_X$ and $f.t\colon X \to Y^I$. Observe that the $p_0$ and $p_1$ are trivial fibrations in $\mathrm{Top}$.
Define $\hat{f}\colon N_f\to Y$ as the composite $f'.p_0\colon N_f\to Y^I\to Y$. Then $f = \tilde{f}.\hat{f}\colon X\to N_f\to X$ is called the 'glueing factorization' of $f$.
The map $\hat{f}$ is always a fibratrion. This is the only appeal to classical algebraic topology.
Now apply the coreflection $k\colon \mathrm{Top}\to \mathrm{Top}_\Delta$ to that diagram.
Then we have:
(1) $k(\hat{f})$ is a fibration in $\mathrm{Top}_\Delta$ and the maps $k(p_0)$ and $k(p_1)$ are trivial fibrations in $\mathrm{Top}_\Delta$ because $k$ preserves fibrations and trivial fibrations.
(2) $k(p')$ is a trivial fibration in $\mathrm{Top}_\Delta$ because it is the pullback (in $\mathrm{Top}_\Delta$) of $k(p_0)$ along $k(f)$. Therefore $k(\tilde{f})$ lies in the smallest localizer by the 2-for-3-property.
Now suppose that $f$ is a weak equivalence in $\mathrm{Top}_\Delta$. Then the same holds for $k(f)$ and the 2-for-3 property yields that $k(f')$ is also a weak equivalence. Consequently $k(\hat{f}) = k(f').k(p_1)$ is a trivial fibration and $k(f) = k(\tilde{f}).k(\hat{f})$ is in the smallest localizer.