As to "why is the unit interval the canonical interval?", there is an interesting universal property of the unit interval given in some observations of Freyd posted at the categories list, characterizing $[0, 1]$ as a terminal coalgebra of a suitable endofunctor on the category of posets with distinct top and bottom elements.
There are various ways of putting it, but for the purposes of this thread, I'll put it this way. Recall that the category of simplicial sets is the classifying topos for the (geometric) theory of intervals, where an interval is a totally ordered set (toset) with distinct top and bottom. (This really comes down to the observation that any interval in this sense is a filtered colimit of finite intervals -- the finitely presentable intervals -- which make up the category $\Delta^{op}$.) Now there is a join $X \vee Y$ on intervals $X$, $Y$ which identifies the top of $X$ with the bottom of $Y$, where the bottom of $X \vee Y$ is identified with the bottom of $X$ and the top of $X \vee Y$ with the top of $Y$. This gives a monoidal product $\vee$ on the category of intervals, hence we have an endofunctor $F(X) = X \vee X$. A coalgebra for the endofunctor $F$ is, by definition, an interval $X$ equipped with an interval map $X \to F(X)$. There is an evident category of coalgebras.
In particular, the unit interval $[0, 1]$ becomes a coalgebra if we identify $[0, 1] \vee [0, 1]$ with $[0, 2]$ and consider the multiplication-by-2 map $[0, 1] \to [0, 2]$ as giving the coalgebra structure.
Theorem: The interval $[0, 1]$ is terminal in the category of coalgebras.
Let's think about this. Given any coalgebra structure $f: X \to X \vee X$, any value $f(x)$ lands either in the "lower" half (the first $X$ in $X \vee X$), the "upper" half (the second $X$ in $X \vee X$), or at the precise spot between them. Thus, you could think of a coalgebra as an automaton where on input $x_0$ there is output of the form $(x_1, h_1)$, where $h_1$ is either upper or lower or between. By iteration, this generates a behavior stream $(x_n, h_n)$. Interpreting upper as 1 and lower as 0, the $h_n$ form a binary expansion to give a number between 0 and 1, and therefore we have an interval map $X \to [0, 1]$ which sends $x_0$ to that number. Of course, should we ever hit $(x_n, between)$, we have a choice to resolve it as either $(bottom_X, upper)$ or $(top_X, lower)$ and continue the stream, but these streams are identified, and this corresponds to the identification of binary expansions
$$.h_1... h_{n-1} 100000... = .h_1... h_{n-1}011111...$$
as real numbers. In this way, we get a unique well-defined interval map $X \to [0, 1]$, so that $[0, 1]$ is the terminal coalgebra.
(Side remark that the coalgebra structure is an isomorphism, as always with terminal coalgebras, and the isomorphism $[0, 1] \vee [0, 1] \to [0, 1]$ is connected with the interpretation of the Thompson group as a group of PL automorphisms $\phi$ of $[0, 1]$ that are monotonic increasing and with discontinuities at dyadic rationals.)
You're spot on with Hatcher's construction. First he constructs the space $EG$ as a $\Delta$-complex, but this can easily be upgraded to a simplicial set $\mathcal{E}G$ such that $EG$ is the geometric realisation of $\mathcal{E}G$:
In fact let $\mathcal{E}G_n:=G^{n+1}$ and define the face maps as $(g_0, ..., g_n) \mapsto (g_0, ..., \widehat{g_i}, ..., g_n)$ and the degeneracy maps as $(g_0, ..., g_{n-1}) \mapsto (g_0, ..., g_i, g_i, ...., g_n)$. Then $G$ acts by simplicial maps via $g(g_0, ..., g_n) \mapsto (gg_0, ..., gg_n)$ and therefore it also acts simplicially on the realisation $EG:=|\mathcal{E}G|$. In fact $G$ acts freely and $EG$ is contractible so that $EG/G$ is a model of $BG$.
He also explains with the bar notation how this space is exactly equal to the geometric realisation of the nerve of $G$ (even if Hatcher does not use these words). Explicitly: $[g_1|g_2|...|g_n]$ is the image in $EG/G$ of all the simplices $g_0\cdot(1,g_1,g_1g_2,...,g_1..g_n)$ in $EG$. The face maps in bar notation are $[g_1|...|g_n]\mapsto [g_2|...|g_n]$ for $i=0$, $[g_1|...|g_n]\mapsto[g_1|...|g_ig_{i+1}|...g_n]$ for $0<i<n$ and $[g_1|...|g_n]\mapsto[g_1|...|g_{n-1}]$ for $i=n$ which are exactly the face maps for the nerve $N(G)$. The degeneracy maps in bar notation are exactly $[g_1|...|g_{n-1}] \mapsto [g_1|...|g_i|1|g_{i+1}|...|g_{n-1}]$ which is also exactly the same as for $N(G)$. Therefore $EG/G = |N(G)|$ (not just a homotopy equivalency, a very canonical homeomorphism)
Best Answer
The correct construction for a topological category is as follows:
If C is a topological category, we can replace it trivially with a simplicial category by taking the simplicial singular complex associated to each hom-space. By abuse of notation, we will call this functor $Sing$.
Now it suffices to give the answer for simplicial categories.
However, to find the classifying space of a simplicial category, we take its associated quasicategory by looking at the homotopy coherent nerve.
The homotopy coherent nerve is usually constructed formally as the adjoint of another functor called $\hat{FU}$, which is the extension of the bar construction $\bar{FU}$ for the associated comonad $FU:Cat\to Cat$ of the free-forgetful adjunction $U:Quiv\rightleftarrows Cat:F$.
Specifically, given any comonad based at $X$, we can form the bar construction, which gives us a functor from $X\to X^{\Delta^{op}}$. This is done by taking objects to be $F_k=F^{k+1}$ the degeneracies to be instances of the comultiplication map $s_i:F_k\to F_{k+1}=F^i\mu F^{k-i}:F^{k+1}\to F^{k+2}$ and faces given by the appropriate application of the counit (the idea is similar to the above, and I leave it as an exercise). In particular, we may take the whole simplicial object in $End(X)$, which gives us our functor $\bar{F}:X\to X^{\Delta^{op}}$
Back to our specific case, we resolve the comonad $FU:Cat\to Cat$ to a functor $\bar{FU}:Cat\to Cat_\Delta$ (since the resolution is trivial on objects , we can say this with a straight face). Restricting $\bar{FU}$ to $\Delta$, which can always be embedded as a full subcategory of $Cat$. By general abstract nonsense, any functor $X\to C$ where C is cocomplete lifts to a unique colimit preserving functor $Psh(X)\to C$ (since taking presheaves gives a "free" cocompletion). Standard notation suggests that we call this functor $\hat{FU}:sSet\to Cat_\Delta$, but following Lurie, we will call it $\mathfrak{C}$. In particular, this functor has a right adjoint called the homtopy coherent nerve, which we can compute as follows:
$$\mathcal{N}(C)_n:=Hom_{Cat_\Delta}(\mathfrak{C}(\Delta^n),C)$$.
for any simplicial category $C$.
Returning to your original case, $BC=\mathcal{N}(Sing(C))$ for a topological category $C$, and for a topological group, we need only notice that a topological group is identical to a one-object Top-enriched category, all of whose morphisms are invertible (something like this).
As for why this is the right definition, I fear I must refer you to Lurie's HTT. It relies on a proof of a certain Quillen equivalence, and alas, the margins are too small...
Edit: Alright, so the reason why they agree is covered in ยง4.2.4 of HTT, I'm pretty sure.