You have the right idea, but it isn't written well. You need to use your observations to write your argument in reverse. Something like this:
Given ca_{n} [itex]\rightarrow[/itex] L we will show that a_{n} [itex]\rightarrow[/itex] L/c, which is a contradiction.
Suppose[itex]\ \epsilon > 0[/itex]. Then there is N > 0 such that:
 ca_{n}  L < c[itex]\epsilon [/itex]
for all n > N. Therefore
ca_{n}  L = c(a_{n}  L/c = c(a_{n}  L/c < c[itex]\epsilon[/itex]
which gives, upon dividing that last inequality by c,  a_{n}  L/c < [itex]\epsilon[/itex]
for all n > N.
