Mariano already answered the question, but let me make two additional remarks:

1. Actually the proof of GAFT is constructive and can be (at least sometimes) used to get explicitly a left adjoint functor. In this case, you can see directly the "big coequalizer" whose set-theoretic existence issue is dealt with the solution set condition: just consider all maps from your space $X$ into Hausdorff spaces which are surjective; this bounds the cardinality of the Hausdorff spaces, and thus up to isomorphism there is only a set of them.

2. Another construction of the left adjoint to $\mathsf{Haus} \to \mathsf{Top}$ works as follows: Let $X$ be a topological space, and consider the equivalence relation $\sim$ generated by: If $x,y$ cannot be separated by disjoint open sets, then $x \sim y$. Then $H(X):=X / \sim$ has the property that every map from $X$ into a Hausdorff space uniquely factors through $X \to H(X)$. If $H(X)$ was Hausdorff, we would be done. But this is not always the case. Instead, we have to repeat this construction: $X \to H(X) \to H(H(X)) \to H(H(H(X))) \to \dotsc$, then take the colimit $H^{\omega}(X)$, and make again $H^{\omega}(X) \to H(H^{\omega}(X)) \to \dotsc ...$. You can continue this for every ordinal number. Since $X$ is a set and all these maps are quotient maps, at some stage we get an isomorphism, which is the desired Hausdorff quotient.

It is interesting when we arrive at this stage, see my question about the nonhausdorff dimension.

Mariano already answered the question, but let me make two additional remarks:

1. Actually the proof of GAFT is constructive and can be (at least sometimes) used to get explicitly a left adjoint functor. In this case, you can see directly the "big coequalizer" whose set-theoretic existence issue is dealt with the solution set condition: just consider all maps from your space $X$ into Hausdorff spaces which are surjective; this bounds the cardinality of the Hausdorff spaces, and thus up to isomorphism there is only a set of them.

2. Another construction of the left adjoint to $\mathsf{Haus} \to \mathsf{Top}$ works as follows: Let $X$ be a topological space, and consider the equivalence relation $\sim$ generated by: If $x,y$ cannot be separated by disjoint open sets, then $x \sim y$. Then $H(X):=X / \sim$ has the property that every map from $X$ into a Hausdorff space uniquely factors through $X \to H(X)$. If $H(X)$ was Hausdorff, we would be done. But this is not always the case. Instead, we have to repeat this construction: $X \to H(X) \to H(H(X)) \to H(H(H(X))) \to \dotsc$, then take the colimit $H^{\omega}(X)$, and make again $H^{\omega}(X) \to H(H^{\omega}(X)) \to \dotsc ...$. You can continue this for every ordinal number. Since $X$ is a set and all these maps are quotient maps, at some stage we get an isomorphism, which is the desired Hausdorff quotient.

It is interesting when we arrive at this stage, see my question about the nonhausdorff dimension.

Mariano already answered the question, but let me make two additional remarks:

1. Actually the proof of GAFT is constructive and can be (at least sometimes) used to get explicitely a left adjoint functor. In this case, you can see directly the "big coequalizer" whose set-theoretic existence issue is dealt with the solution set condition: just consider all maps into Hausdorff spaces which are surjective; this bounds the cardinality of the Hausdorff spaces, and thus there is only a set of isomorphism-classes of them.

2. Another construction of the left adjoint to (Haus) -> (Top) works as follows: Let $X$ be a topological space, and consider the equivalence relation $\sim$ generated by: If $x,y$ cannot be separated by disjoint open sets, then $x \sim y$. Then $H(X):=X / \sim$ has the property that every map from $X$ into a Hausdorff space uniquely factors through $X \to H(X)$. If $H(X)$ was Hausdorff, we would be done. But this is not always the case. Instead, we have to repeat this construction: $X \to H(X) \to H(H(X)) \to H(H(H(X))) \to \dotsc$, then take the colimit $H^{\omega}(X)$, and make again $H^{\omega}(X) \to H(H^{\omega}(X)) \to \dotsc ...$. You can continue this for every ordinal number. Since $X$ is a set and all these maps are quotient maps, at some stage we get an isomorphism and then the desired Hausdorff quotient.

It is interesting when we arrive at this stage, see my question about the nonhausdorff dimension.

Mariano already answered the question, but let me make two additional remarks:

1. Actually the proof of GAFT is constructive and can be (at least sometimes) used to get explicitly a left adjoint functor. In this case, you can see directly the "big coequalizer" whose set-theoretic existence issue is dealt with the solution set condition: just consider all maps from your space $X$ into Hausdorff spaces which are surjective; this bounds the cardinality of the Hausdorff spaces, and thus up to isomorphism there is only a set of them.

2. Another construction of the left adjoint to $\mathsf{Haus} \to \mathsf{Top}$ works as follows: Let $X$ be a topological space, and consider the equivalence relation $\sim$ generated by: If $x,y$ cannot be separated by disjoint open sets, then $x \sim y$. Then $H(X):=X / \sim$ has the property that every map from $X$ into a Hausdorff space uniquely factors through $X \to H(X)$. If $H(X)$ was Hausdorff, we would be done. But this is not always the case. Instead, we have to repeat this construction: $X \to H(X) \to H(H(X)) \to H(H(H(X))) \to \dotsc$, then take the colimit $H^{\omega}(X)$, and make again $H^{\omega}(X) \to H(H^{\omega}(X)) \to \dotsc ...$. You can continue this for every ordinal number. Since $X$ is a set and all these maps are quotient maps, at some stage we get an isomorphism, which is the desired Hausdorff quotient.

It is interesting when we arrive at this stage, see my question about the nonhausdorff dimension.