A surprisingly effective way to construct counterexamples in model category theory is to just write down all the objects and morphisms involved and try to give the resulting (finite!) diagram the structure of a model category.

Here, we know that a counterexample must fail to be left proper, so start with a diagram$\require{AMScd}$
$$
\begin{CD}
a @>\sim>> b\\
@VVV @VVV\\
c @>>> d
\end{CD}
$$
in which $a \to b$ is a weak equivalence, $a \to c$ is a cofibration, but $c \to d$ is not a weak equivalence. Then $a \to c$ also cannot be a weak equivalence (otherwise $b \to d$ would be one too). Since $a \to c$ and $c \to d$ are not weak equivalences, they must be both cofibrations and fibrations and therefore the same is true of $a \to d$. Then $a \to d$ cannot be a weak equivalence (or it would be an isomorphism), so $b \to d$ is also not a weak equivalence, and therefore is a fibration too. In summary, all the maps are fibrations and $a \to c$, $b \to d$, $c \to d$ are cofibrations while $a \to b$ is a weak equivalence. One can check that this does in fact yield a model category structure (probably the easiest way is to verify that the (acyclic) cofibrations/fibrations are closed under composition and pushout/pullback, and that the factorization axioms hold).

Now, let's try to form the left Bousfield localization at the map $a \to c$, which is already a cofibration between cofibrant objects. All objects are fibrant in the original structure, and the local objects are the ones which have the same maps from $a$ and from $c$, which are the objects $c$ and $d$. The map $c \to d$ was not a weak equivalence originally, so it has to still not be one in the localization. However, making $a \to c$ a weak equivalence also makes $b \to d$ a weak equivalence because it is the pushout of the acyclic cofibration $a \to c$, which contradicts two-out-of-three.

simplicialmodel category. Is the "simplicial" condition removed somewhere? $\endgroup$ – Tim Campion Mar 13 at 22:03