domain of initial $f : X \rightarrow Y$ in Haus equipped with coarsest topology? If $f:X\rightarrow Y$ is initial in category Top then
it is easy to proof that 

(!) the topology on $X$ is the set of preimages of open sets in $Y$. 

Just construct topology $Z$ having
the same underlying subset as $X$ and let the set of these preimages
serve as topology on it. Then from $g:Z\rightarrow X$ with $x\mapsto x$
it is clear that $fg$ is continuous so the conclusion that $g$ is
continuous can be made. Then we are ready. 
But now my question: 

what if we do not work in $\textbf{Top}$ but in category $\textbf{Haus}$?

The constructed topology $Z$ does not have to be a Hausdorff space (or am I overlooking something here?) and if the fact that $f$ is initial in $\textbf{Haus}$ would work then it would justify the conclusion that $g$ can be recognized as an arrow in $\textbf{Haus}$. 

Is there a way out? Or - even stronger - is statement (!) not true in $\textbf{Haus}$?

 A: Initial topologies need not be Hausdorff. For example, for any set $X$, the initial topology induced by the unique map $X \to 1$ is Hausdorff if and only if $X$ has cardinality $\le 1$.
Worse, there need not be an initial topology among the Hausdorff ones. For example, suppose $X$ is an infinite set with the same cardinality as $\mathbb{R}$. Then, for every subset $S \subseteq X$, if $S \ne X$ and $S \ne \emptyset$, then there exists a Hausdorff topology on $X$ such that $S$ is not open. (If $S$ has cardinality strictly less than $X$, then topologise $X$ so that $X \cong \mathbb{R}$, and observe that every non-empty open subset of $\mathbb{R}$ has cardinality equal to $\mathbb{R}$; otherwise, choose a suitable bijection $X \cong \mathbb{R}$ so that $S$ is identified with $[0, 1] \subseteq \mathbb{R}$, which is not open.) Hence, the only topology on $X$ that is coarser than all the Hausdorff topologies on $X$ is the indiscrete topology on $X$, which is not Hausdorff.
A: A continuous $f: X \to Y$ is by definition initial iff [$(∀Z\in\mathbf{Top} ) (∀g: Z \to X)$ $g$ is continuous iff $f g$ is continuous]. In $\mathbf{Top}$ that's equivalent to $X$ having the initial topology induced by $f$. And it is also equivalent in $\mathbf{Haus}$ since (as you noted) initial morphisms in $\mathbf{Haus}$ are precisely the embeddings.
Let $f: X \to Y$, $Y$ Hausdorff. Then the initial topology on $X$ is Hausdorff if and only if $f$ is injective. If $f$ is not injective then initial topology on $X$ isn't even $T_0$ since points with the same image are not topologically distinguishable. On the other hand if $f$ is injective, then initial topology on $X$ is subspace topology and hence Hausdorff.
Let $f: X \to Y$ be initial in $\mathbf{Haus}$ (so $X$, $Y$ are Hausdorff). Then $f$ is injective.
Proof: Let $Z$ be some Hausdorff space, $∅ ≠ A ⊊ Z$ non-open. Let $f(x) = f(x')$, then let $g: Z \to X$ map $A$ to $x$ and $Z \setminus A$ to $x'$. Then $g$ is not continuous since $A$ is non-open and $X$ is Hausdorff. But $fg$ is constant so $f$ cannot be initial.
So if $f: X \to Y$ is initial in $\mathbf{Haus}$ then the initial topology on $X$ is Hausdorff by the proposition above and so $f$ is an embedding.
Note that the same proofs work for $T_0$, $T_1$, $T_3$, $T_{3\frac{1}{2}}$ and any full subcategory of $T_0$ spaces closed under subspaces and containing some non-discrete space.
