Question: I am having difficulty with $\vdash (p\rightarrow q)\vee(q\rightarrow r)$.
My idea to solve this was to first assume p, somehow find q, write that $p\rightarrow q$ and then write $\vee i$ for the second part of the statement. I know that just assuming p is not enough to solve this, so I need to make one of two more assumptions about q and r. If I just assume q to be true, then I will get that $q \rightarrow q $, which I don't want. Below is the best I could do.

$ 1. \space p$ assumption 

$ 2. \space q$ assumption 

$ 3. \space q \rightarrow q \rightarrow i \space 2$

$ 4. \space p \rightarrow (q \rightarrow q) \space \rightarrow i \space 1, \space 2-3$

$ 5. \space (p \rightarrow (q \rightarrow q)) \space \vee (q \rightarrow r) \space \vee i \space 4 $

Answer: 
Sorry, but your proof is getting nowhere ... Always focus on the *main* connective of what you are trying to prove. In this case, that is a $\lor$. So, even though there are $\rightarrow$'s appearing in the formula, since they are not the main connective, you should not try to do a conditional proof.

HINT

When what you are trying to prove is a disjunction (i.e when the main conective is a $\lor$), there are typically 3 strategies:

1. If you are lucky, you may be able to get to one of the disjuncts .. no such luck here!
2. If you have a disjunction to work with, then set up a proof by cases ($\lor$ Elim) on that disjunction ... unless you have the Law of Excluded Middle to work with, you have no disjunction to work with!
3. Try a proof by contradiction. ... Yes, do this by a proof by contradiction!
$\neg P\vee Q$ is the same as $P\to Q$.

$(p\to q)\vee (q\to r)\equiv(\neg p\vee q)\vee(\neg q\vee r)\equiv\neg p\vee(q\vee \neg q)\vee r\equiv \neg p\vee T\vee r\equiv T$.

This is using the Law of the Excluded Middle ($q\vee \neg q\equiv T$).

Without knowing what kind of deduction system you are using, this may or may not count as a proof.