It's not true that the homotopy colimit is the colimit in the homotopy category. In fact, colimits (other than disjoint unions, say) tend not to exist there.
Here's a bit of intuition for homotopy colimits. (A nice reference is Dan Dugger's primer on the subject.) Suppose you have a diagram of, say spaces or simplicial sets, as in
$$\begin{array}{rcl}
A&\to& B\\
\downarrow&&\downarrow\\
C&\to&
\end{array}$$
To form the push-out in an ordinary sense, you would take the disjoint union $B \sqcup C$, and then quotient by the relation that for each $a \in A$, the image of $a$ in $B$ is identified with that in $C$. This, however, is not homotopy invariant, because the operation of quotienting in this way is not well-behaved homotopically. A better variant (from this point of view, at least) is not to quotient by this relation, but to glue in paths for it. That is, if $i: A \to B, j: A \to C$, then to form the homotopy push-out of this diagram, you should draw in paths from $i(a)$ to $j(a)$ for each $a \in A$.
This is one explicit construction of the homotopy push-out which works for spaces that are nice (say, CW complexes) or for arbitrary simplicial sets (for the more general theory, you should look up the Bousfield-Kan formula). You can check that in this case the operation just described is homotopically well-behaved, and that it is weakly equivalent to the usual colimit for projectively cofibrant diagrams.
So, let's take this as our explicit construction of the homotopy push-out of spaces or simplicial sets. Now, we want to describe maps from the homotopy push-out, which I'll call $D$, into an arbitrary space $X$. By definition, this is given by maps $B \to X$, $C \to X$, and a homotopy $A \times I \to D$ of the restriction $A \to B \to X$ with $A \to C \to X$. This follows from the other construction as $D$ by gluing in lots of paths.
Note that the homotopy itself is part of the data; it's not enough to just give two maps that happen to be homotopic. We can think of this as a space of maps, and its connected components are the morphisms $[D, X]$ in the homotopy category.
Now, if the space of maps $D \to X$ were just the space of maps $B \to X, C \to X$ whose restriction to $A$ were homotopic, then you're right: $D$ would be the push-out in the homotopy category. But the problem is that the homotopy has to be included in the data. Here's an explicit example to illustrate this. Let $B = C = \ast$. Then the homotopy push-out is given by drawing paths $\ast \to \ast$ for each $a \in A$; this, in other words, is the suspension of $A$ (whether it's reduced or unreduced depends on whether you work in the pointed or unpointed category). What the above says is that to give a map $\Sigma A \to D$ is to give two points of $D$, and a homotopy between the two constant maps $A \to D$ given by these points; this is reasonable. More to the point, in this case the homotopy push-out is definitely not the push-out in the homotopy category, because $\ast$ is final in the homotopy category and that push-out would be $\ast$.
Finally, note that even homotopy inverse limits are not inverse limits in the homotopy category (e.g. because of the Milnor exact sequences with $\lim^1$ terms).
A co-retraction is defined as a map having a left inverse. We know that $\mu : M(i) \to X \times \mathbb{I}$ is co-retraction with retract $\rho$. Therefore $1_B \times \mu : B \times M(i) \to B \times X \times \mathbb{I}$ is a co-retraction with retract $1_B \times \rho$. This shows that also $\mu' : M(1_B \times i) \to B \times X \times \mathbb{I}$ is a co-retraction.
If you accept that the second diagram on p. 268 is a pushout diagram, then also the first diagram ($\ast$) is one simply because you have homeomorphisms between the four spaces at the corners of the two squares producing a commutative cubical diagram.
By the way, $1_B \times i$ is always a cofibration if $i$ is one. The general proof is much harder than the proof based on the assumptions in your question.
Added:
The natural map $h : M(1_B \times i) \to B \times M(i)$ is constructed via the universal property of the pushout.
By definition, $M(1_B \times i)$ occurs in lower right corner of a pushout diagram - that we denote for the moment by ($\ast\ast$) - based on the two maps $B \times A \times 0 \to B \times X \times 0$ and $B \times A \times 0 \to B \times A \times \mathbb{I}$ of diagram ($\ast$).
Similarly $M(i)$ occurs in a pushout diagram based on the two maps $A \times 0 \to X \times 0$ and $A \times 0 \to A \times \mathbb{I}$. Multiplying this diagram from the left (Cartesian product!) with $B$ produces the commutative diagram ($\ast)$. The universal property of the pushout gives you a unique map $h : M(1_B \times i) \to B \times M(I)$ with the well-known properties.
Ronnie Brown proves that ($\ast)$ is pushout diagram. Since both ($\ast\ast$) and ($\ast$) are pushout diagrams, the map $h$ must be a homeomorphism.
Best Answer
Well, if you have handled the finite case, now you are left with "taking the colimit", exactly as you mentioned!
More generally, suppose that $f_k: Y \to Z_k$ is a family of cofibrations cofibration, where $\{Z_k\}_{k \in \mathbb{N}}$ is a diagram $Z_1 \to Z_2 \to Z_3 \to \ldots$ made of cofibrations. Suppose also that $f_k$ are consistent with maps $Z_k \to Z_{k+1}$,and set $Z = \textrm{colim} Z_k$. Then $f = (Z_1 \to Z) f_1$ is a cofibration. You will use $Y=X_n$ and $Z_k = X_{n+k}$ to recover the result in your case. Note that the hypothesis holds true because of what you showed in the finite case.
The proof is very general and holds for any model category, if you know what they are. Indeed, if you want to know whether a map is a cofibration, you have to check if it has the left lifting property against all fibrations. I suggest you make diagrams yourself to follow the proof.
Let $U \to V$ be a fibration, and let $Y \to U, Z \to V$ maps so that the square $Y, U, Z, V$ commutes. We want to know if there exist a diagonal map - a "lifting" - from $Z$ to $U$ that makes the two triangles commute. Restrict the square to $Y, Z_1, U, V$. Since $Y \to Z_1$ is a cofibration, you can find a map $Z_1 \to U$ that makes the triangles commute. Now construct a square $Z_1, U, Z_2, V$. Since $Z_1 \to Z_2$ is a cofibration, you can extend the map $Z_1 \to U$ to a map $Z_2 \to U$. Continuing in this fashion, you get maps $Z_k \to U$ that commutes with the diagram maps $Z_n \to Z_{n+1}$ and that correctly restrict to $Y$. Taking the colimit of such maps you get your desired lifting $Z \to U$, and you are done.