Problem: A matrix $A$ is not invertible iff $0$ is an eigenvalue of $A$
Suppose $\det(A)=0$ ($A$ is not invertible).
Then $\det(A-\lambda I)=0$ iff $\lambda =0$.
$\therefore A$ is not invertible iff $\lambda = 0$.
I was wondering if I'm using valid logical induction steps to prove this.
It is not valid as you wrote it, but rearranging the same idea could make it correct.
You wrote "$\det(A-\lambda I)=0$ iff $\lambda =0$". That is not true in general, and is equivalent to the statement that $0$ is the only eigenvalue of $A$. You conclude "$A$ is not invertible iff $\lambda = 0$," which is not a clear statement as you haven't quantified $\lambda$.
However, it is true that $A$ is not invertible if and only if $\det(A) = 0$ if and only if $\det(A-0 I)=0$ if and only if $0$ is an eigenvalue for $A$. Therefore, $A$ is not invertible if and only if $0$ is an eigenvalue for $A$. That is valid as long as you are allowed to cite theorems that give you each of the equivalencies above.
Alternatively, perhaps you have a theorem that tells you that a matrix is invertible if and only if its nullspace is trivial. Having nontrivial nullspace is equivalent to having $0$ as an eigenvalue, directly for the definitions of what those words mean: both mean there exists $v \neq 0$ such that $Av=0$.