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.