This is my own proof, I just need to know if I am doing this correctly.
(By the way, I'm 100% sure this is not an original proof, I just came with it by myself for a homework question and want to make sure it is right before turning it in).
Problem: Let $R$ be an integral domain, then the ring of polynomials $R[x]$ with coefficients in $R$ is an integral domain.
My proof:
Let $f,g \in R[x]$, such that $\mathrm{deg}(f)=m$ and $\mathrm{deg}(g)=n$ Therefore we can represent $f$ and $g$ by $$f = \sum_i^m a_i x^i$$ $$g = \sum_j^n b_j x^j$$
where $a_i, b_j \in R \quad \forall 0 \leq i \leq m, 0 \leq j \leq m$
We know that $R[x]$ is an integral domain if and only if $R[x] \setminus\{0_{R[x]}\}$ is closed under multiplication.
Let $f \neq 0 \neq g$. Therefore we are guaranteed that, of all the coefficients in $f$ and $g$, at least $a_m \neq 0$ and $b_n \neq 0$.
We can multiply polynomials in the following way
$$fg = \bigg(\sum_i^m a_i x^i\bigg) \bigg(\sum_j^n b_j x^j\bigg) = \sum_i^m \sum_j^n (a_ib_j)x^{i+j}$$
By what was shown before, we are guaranteed that, out of all the elements of this double sum, at least $(a_mb_n)^{m+n} \neq 0$. Therefore, $fg \in R[x] \setminus\{0_{R[x]}\}$ which means that $R[x] \setminus\{0_{R[x]}\}$ is closed under multiplication. Therefore, $R[x]$ is an integral domain.
Since the comments cleaned up your essentially correct proof, I submit this as a congratulations on your proof and to remove this from the unanswered questions!