Question: I am given that
$$a \equiv 11 \pmod {19}$$
and $$ c \equiv 13a \pmod {19}$$ 
and I am asked to solve for a $c$ that is between $0$ and $18$ inclusive.

From looking at problems online, it seems that the easiest way to do this is as follows:

$$ c \equiv 13(11) \pmod {19}$$
$$ c \equiv 143 \pmod {19}$$
$$ 143 = 19(7) + 10$$
therefore
$$ c = 10$$
because $$ 19 \mid (143-10). $$

So I have two questions: how is it possible to substitute $11$ as $a$, as my book does not prove this nor mention it and I have a very hard time with proving things myself?

And is that last step (solving for $c$) a trial and error approach, as it seems to be that way?

Answer: 
Why does your book proceed this way? That's because congruences, like equalities, are compatible with addition and multiplication, i.e.
$$a\equiv a',\quad b\equiv b'\pmod n\implies\begin{cases} a+b\equiv a'+b',\\ab\equiv a'b'\pmod n
\end{cases}$$
As to the explicit value of $c$, it is not at all obtaind by trial and error, but by a plain old *euclidean division* of $143$ by $19$.
$a\equiv 11\pmod{19}$ means that $19\, |\, (a-11)$.   

Note that it implies $19\, |\, (a-11)13\, =13a-11\cdot13$, which translates back to $13a\equiv 11\cdot13$.

In addition, $11\cdot13= 147$ gives *remainder* $10$ when dividing by $19$, so we will have $c= 13a\equiv147\equiv10\pmod{19}$.

Also note that sometimes calculating by *negative values* can yield shorter calculations. In your case $11\equiv -8$ and $13\equiv -6$, so they can be exchanged $\pmod{19}$, use the above argument (if not convinced yet), and they of course give the same result: $(-8)\cdot(-6)=48=2\cdot19+10\equiv 10\pmod{19}$.