I will proceed with a proof by contradiction.
If we assume that $7$ is reducible, then there exists some $q$ and $r$ in $A$ such that $q$ and $r$ are not units, and $7 = qr$. Using the norm ( $N(a+b\sqrt{-14})=a^2+14b^2$ ), and that fact that an element in a quadratic integer ring is a unit if and only if the norm is equal to $+1$ or $-1$. Also the norm is multiplicative, so we would have that $$N(7) = 49 = N(qr)= N(q)N(r)$$
By considering the factorization of 49 in $\mathbb{Z}$, (1,7,49) are factors fo 49, then $N(q)=N(r)=\pm 7$. If not then $q$ or $r$ would be a unit. So assume that $q = c + d\sqrt{-14}$, then $N(q)=c^2 + 14d^2=7$ for some $c$ and $d$ in $\mathbb{Z}$. This implies that $c^2=7$, which has no solution in $\mathbb{Z}$.
So, 7 must be irreducible in $A$.
My question is if this proof is complete. Looks good to me but I am just a beginner. Thanks.