If gcd$(a,b) = 1$, then I want to prove that $\forall c \in \mathbb{Z}$, $ax + by = c$ has a solution in integers $x$ and $y$. If gcd$(a,b) = 1$, then I want to prove that $\forall c \in \mathbb{Z}$, $ax + by = c$ has a solution in integers $x$ and $y$. 

This is what I was attempting or trying:

Let $d =$ gcd$(a,b)$.

$d|a \Rightarrow \exists k \in \mathbb{Z}$ s.t. $a = dk$.

$d|b \Rightarrow \exists l \in \mathbb{Z}$ s.t. $b = dl$

Substituting these in we get $dkx + dly = c \Rightarrow d(kx + ly) = c \Rightarrow d|c$ and $(kx+ly)|c$.

I'm not sure if this proves it or not. The help would be appreciated!
You have proved that if $d\mid a$ and $d\mid b$, and $c=ax+by$, then $d\mid c$. That is not at all what we want to show.

The result will be easy once we prove:

Lemma: If $\gcd(a,b)=1$, there exist integers $s$ and $t$ such that $as+bt=1$.

Taking the Lemma for granted temporarily, note that if $as+bt=1$ then $a(cs)+b(ct)=c$, so we can take $x=cs$ and $y=ct$ to prove our result. But the hard part is 

Proof of Lemma: We sketch the standard proof. Let $K$ be the set of all positive integers that  can be represented in the form $au+bv$, where $u$ and $v$ range over the integers. The set $K$ is non-empty, so has a smallest positive element $k$.  So there is a $u_1,v_1$ such that $k=au_1+bv_1$. We show that $k=1$. That will complete the proof.

Suppose to the contrary that $k\gt 1$. The number $k$ cannot divide both of $a$ and $b$, for if it did we would violate $\gcd(a,b)=1$. Without loss of generality we may assume $k$ does not divide $a$. Then $a=qk+r$, for some integers $q$ and $r$ such that $0\lt r\lt k$.

But then
$$r=a-qk=a-q(au_1+bv_1)=a(1-qu_1)+b(-qv_1).$$
Since $0\lt r\lt k$, this contradicts the fact that $k$ is the smallest positive integer of the shape $au+bv$.
If $a$ and $b$ are relative primes, the numbers $ax\bmod b$ with $x$ from $0$ to $b-1$ are all different. [Otherwise, $ax\bmod b=ax'\bmod b\implies a(x-x')\bmod b=0$  so that $b$ divides $x-x'$ ($a$ and $b$ having no common factor), a contradiction.] As they are different, they take all values from $0$ to $b-1$, in disorder.

So there is certainly an $x$ such that $ax\bmod b=c\bmod b$, which we can rewrite $ax+by=c$.

For example, $31$ and $17$ are relatively primes. The sequence of $31x\bmod 17$ is
$$0,14,11,8,5,2,16,13,10,7,4,1,15,12,9,6,3.$$

To solve $31x+17y=100$, lookup $100\bmod17=15$, which is $31\cdot12\bmod 17$, and deduce $31\cdot12-17\cdot16=100$.