On the 1-simplex not being fibrant: How can I see this directly? The standard 1-simplex is not a Kan complex, because the homotopy relation on maps to the 1-simplex is not an equivalence relation (which is a property of Kan complexes):
We have the standard 1-simplex $Y := \operatorname{Hom}_{\Delta}( \cdot, [1])$.  There is a subsimplicial set, denoted $\mathbf{1}$, given by the homomorphism $\in \operatorname{Hom}_{\Delta}( \cdot , [1])$ whose image is $1 \in [1]$.
Similarly there is a subsimplicial set denoted $\mathbf{0}$.
Let $d_0$ be the standard map from $[0] \to [1]$  that misses 0. Let $i_0$ be the map (of simplicial sets) $i_0: \operatorname{Hom}(*,[0]) \to \operatorname{Hom}(*,[1])$ induced by $d_1:[0] \to [1]$ and let $i_1$ be a similar map induced by $d_0$.
Then there is a homotopy from $i_0$ to $i_1$ but there is no homotopy $h$ from $i_1$ to $i_0$:  There is no degree 1 map $h: \operatorname{Hom}(*,[0]) \to Y$ such that $d_0 h=i_0$ and $d_1h=i_1$; in fact there are only $3$ candidates for $h(\operatorname{Id}_{[0]})$ and all of them fail.

Now I want to see that $Y$ is not a Kan complex directly.  This is true because there are two nondegenerate mappings (mapping one of the 1 simplices nondegenerately) of the horn $\Lambda_0^2$ into $Y$ and there are no nondegenerate mappings of $\Delta^2 \to Y$.  By the simplicial identities, $d_i$ of any degenerate simplex is degenerate.
After typing this out, I answered my own question above, unfortunately.  But if you have anything to add please do.
 A: Notation: I'm calling the $n$-simplex $\Delta^n$.
Standard introductions to weak Kan complexes (which the 1-simplex certainly is) would make you believe that the 1-simplex has fillers for inner horns, i.e. $\Lambda^n_i$ where $0 < i < n$, but not necessarily for outer horns, i.e. $\Lambda^n_0$ and $\Lambda^n_n$. Filling outer horns should mean being able to invert a morphism/1-simplex.
So let us search an outer horn whose filling would witness invertibility of a non-invertible morphism, and then show that it can't be filled! The obvious non-invertible morphism is the unique non-degenerate 1-simplex, i.e. the identity map, which I'll call $f\colon \Delta^1 \to \Delta^1$.
The horn $\Lambda^2_0$ whose filling would expose a (one-sided) inverse for this morphism would consist of a wedge of two morphisms, $f$ and your $i_0$. (The order is important here.) Filling this horn would mean to find a morphism/1-simplex $g$ such that $g \circ f \simeq i_0$. But $g$ would have to go from $\mathbf{1}$ to $\mathbf{0}$, and there is no such morphism! So there can't be a filler for this horn, since it wouldn't even have a boundary.
