minimal embeddings of topological spaces into connected spaces Defintions: 
Let $X$ be a topological space.
1) A connected space $Y$ is a minimal connected ambient (m.c.a for short) space for $X$ if there exists an embedding $i:X\mapsto Y$, and for every connected space $Y'$ into which $X$ can be embedded, there exists an embedding $j:Y\mapsto Y'$.
2) A connected space $Y$ is a smallest connected ambient (s.c.a for short) space for $X$ if there exists an embedding $i:X\mapsto Y$, and for every connected space $Y'$ into which $X$ can be embedded, and for any embedding $j:X\mapsto Y'$ it holds that $|Y-i(X)|\leq |Y'-j(X)|$.
 (That is we added the least amount of points to $X$ to create $Y$).
Remarks:
1) If $X$ is connected then clearly $X$ itslef is both a minimal and smallest connected ambient space.
2) As noted by John, every topological space $X$ can be embedded into it's cone $CX$, which is connected.
This raises some quetions:
Assume $X$ is not connected.
1) Is it always possible to embedd $X$ in a smallest (s.c.a) connected ambient space?
Udate: The answer is yes. (According to the construction given by Kaj Hansen, which is the open extension). Since we assume $X$ is not connected, any connected space in which $X$ embedds must contain points that are not in the image of $X$. So it must contain at least one more point. Now this trivial lower bound is achieved by the wonderful construction of Hansen).
2) Is it always possible to embedd $X$ in a minimal (m.c.a) connected ambient space? I suspect the answer is nt, and that for most spaces $X$ an there will not be an m.c.a.
3) When the m.c.a (s.c.a) exists, are they unique? (=are they homeomorphic? or even hoemomorphic by a homeomorphism which takes the copy of $X$ in one of them into the copy in the other?)
4) If $Y$ is an m.c.a for $X$ is it also an s.c.a for it? (What about the other direction?)
5) What happens when we add additional structue? (in particular order structure).
Let $(X,<_X)$ be an ordered topological space.
Say $(Y,<_Y)$ is a minimal ordered connected ambient (m.o.c.a for short) space for $X$ if there exists an order-preserving embedding $i:(X,<_X)\mapsto (Y,<_Y)$, and for every ordered connected space $(Y',<_Y')$ into which $(X,<)$ can be embedded , there exists an (order-preserving) embedding $j:(Y,<_Y)\mapsto (Y',<_Y')$.
We define similarly (see def2 above) a smallest ordered connected ambient (s.o.c.a for short) space for $X$
Now we can ask: are there always s.o.c.a, m.o.c.a?
Example:
$\mathbb{Q}$ is totally disconnected. (In the order topology). However it can be embedded in $\mathbb{R}$ which is connected. (This is because the order topology in $\mathbb{Q}$ and the subspace topology on it inherited from $\mathbb{R}$ coincide).
$\mathbb{R}$ is both an m.o.c.a and an s.o.c.a for $\mathbb{Q}$. This follows from the fact that every ordered topological space which is connected must satisfy the completeness axiom about existence of suprema, and hence if it contains $(\mathbb{Q},<_\mathbb{Q})$ it must contain a copy of $(\mathbb{R},<_\mathbb{R})$.
Note that if we do not require the ambient space to be ordered, there is a smaller suitable ambient space (the open extension). In other words the s.c.a and s.o.c.a are differnt in this case.
 A: Here is an idea that I think might be worth your consideration, though it does not take an ordering into account.
Consider a set $X$ endowed with the discrete topology: a"worst case scenario" as far as connectedness is concerned.  Add a single point to $X$—call it $\delta$—and define the following topology on $X \cup \{\delta \}$:

*

*If a subset $U \subset X \cup \{ \delta \}$ does not contain $\delta$, then it is open, as it would be in $X$ under the discrete topology.


*If a subset $U \subset X \cup \{ \delta \}$ contains $\delta$, then $U$ is open $\iff U = X \cup \{ \delta \}$.
Notice that we have preserved the original topology on $X$, so there is a natural embedding $X \hookrightarrow X \cup \{ \delta \}$.  Moreover, the only open set containing $\delta$ is the whole space!  Thus, $X \cup \{ \delta \}$ cannot be expressed as the union of two nonempty, disjoint open sets.  In other words, $X \cup \{ \delta \}$ is connected.  Note also that this works regardless of the original topology on $X$: replace the first bullet with "If $\delta \notin U$, then $U$ is open $\iff$ it is open in $X$" .
So not only is this a one-point compactification of $X$; it's also a one-point connectification.

P.S. this general idea has been very kind to me in the past, so it might be worth bearing in mind.  See here, for instance, where it can be used to show that the intersection of two compact sets need not be compact.
A: If the totally disconnected space is second countable then, since it is normal, it can be embedded in R^(infinity), a metrisable connected space.
