Is there a canonical way to connect a topological space? It is known that each topological space $X$ admits a Hausdorffization - which means that every topological space can be "approximated" by the corresponding unique Hausdorff space for which each continuous mapping from $X$ to a Hausdorff space factors uniquely through the Hausdorffization.
For given topological space is it possible to connect it in a canonical way, i.e. to find a corresponding connected space with the property as above?
 A: As far as I can tell, no. 
If you mean canonical to mean a left adjoint to the full inclusion of categories, in the sense that it is dual to $\mathrm{Conn} \hookrightarrow \mathrm{Top}$, then it has to preserve coproducts, which cannot occur by  Qiaochu's answer here.
The idea is to consider a two point space, which is the coproduct of two one point spaces in $\mathrm{Top}$, but there is no coproduct in the category of connected spaces.
You can see here for an elaboration for the case of path connected spaces.
If you google "connectification of topological spaces" some papers show up discussing various constructions.
A: No, it is not true. Consider $X = \{0,1\}$ a set with two elements, endowed with the discrete topology. Suppose there were a connected space $X'$ and a continuous map $i : X \to X'$ such that for all connected spaces $Y$ and for all continuous maps $f : X \to Y$, there exists a unique map $f' : X' \to Y$ such that $f' \circ i = f$.
Then first I want to show that $i$ is injective. For this consider $f : X \to [0,1]$ the inclusion. Then there exists a unique map $f' : X' \to [0,1]$ such that $f' \circ i = f$. But $f$ is injective, therefore so is $f'$ (consider that if $i(0) = i(1)$, then $f(i(0)) = f(i(1)) \implies 0 = 1$).
Next I want to show that $i$ is surjective. Let $Y = \{a,b\}$ be a set with two elements, endowed with the trivial topology (the only open sets are $\varnothing$ and $Y$). Consider $f : X \to Y$, $f(0) = a$, $f(1) = b$. Then $f$ is continuous, hence there exists a unique map $f' : X' \to Y$ such that $f' \circ i = f$.
Now suppose to the contrary that $i$ weren't surjective. Then I can find two different continuous maps $f' : X' \to Y$ such that $f' \circ i = f$. Either I can send all the elements of $X' \setminus \{i(0), i(1)\}$ to $a$, or I can send them to $b$, while in both cases I map $i(0) \mapsto a$ and $i(1) \mapsto b$. That's a contradiction, hence $i$ is surjective.
It follows that $i$ is bijective, and $X'$ has two elements, $i(0)$ and $i(1)$. But if I consider again $f : X \to [0,1]$ the inclusion, then I find that $i(0)$ and $i(1)$ must be separated by open sets (the preimages of, say, $[0,1/4)$ and $(3/4,1]$ by the induced map $f' : X' \to [0,1]$). Hence $X'$ is actually discrete, which is contradictory with it being connected.
