Question about the proof of: If $m \lt n \Rightarrow f : S^m \rightarrow S^n$ is nullhomotopic. 
I don't understand the red-underlined sentence.  What are they saying is contractible?  Are they saying $\text{im}|\phi|$ is contractible or $|L| - \{\text{point}\}$ is contractible?  And how do either of these being contractible imply that $|\phi|$ is nullhomotopic?
 A: They say that $|L|-$ point is contractible. This is equivalent to saying that there is a continuous map $H:[0,1]\times |L|-point\rightarrow |L|-point$ such that $H(0,x)=x, H(1,x)=p$. You can define $\phi_t(x)=H(t,\phi(x))$ which is an homotopy between a constant map and $\phi$.
A: $|\varphi|$ is not surjective, so it misses at least one point of $|L|$.  Call that point “point”.  The image is contained in $|L|$ with that point deleted.  That space ($L'=L \setminus \left\{\text{point}\right\}$, not $L$ itself) is contractible.  Contractible means there is a point $p \in L'$ and homotopy $H:[0,1] \times L' \to L'$, such that $H_1$ is the constant map $L' \to \left\{p\right\}$.  The map $\tilde H(t,x) = H(t,|\varphi|(x))$ is a homotopy between $|\varphi|$ and a constant map.  
You seem to have some very specific questions like: why is the $m$-skeleton of an $(m+1)$-simplex, with a point deleted, contractible?  That may already be an exercise in the book you're reading.  Whether or not it is, it might be a good idea to try to prove that, and ask a different question if you run into trouble.
