Let $\mathcal{P}(X_N)$ be the power set of a set $X$ with $N$ elements. I am trying to prove by induction that its cardinality $\mid \mathcal{P}(X_N) \mid = 2^N$. Firstly, I think it helps to observe that this can pretty much be "rewritten" or rather fleshed out as
$$\mid \mathcal{P}(X_N) \mid = 2^N = 2^{N-1} \cdot 2^1 = 2 \cdot \mid \mathcal{P}(X_{N-1}) \mid.$$
See that if $N = 0$, $X$ is then the empty set $\emptyset$ and {$\emptyset$} is its only subset; thus, $\mathcal{P}(X_0)$ = $2^0 = 1$, which is correct.
(*) So assume $\mid \mathcal{P}(X_N) \mid$ does in fact equal $2^N$. Then we see that
$$\mid \mathcal{P}(X_{N+1}) \mid = 2^{N+1} = 2^N \cdot 2^1 = 2 \mid \mathcal{P}(X_N) \mid,$$
which is true by (*), our inductive hypothesis. My question is (as you might expect if you have read any of my other posts)...is this a correct use of induction? I apologize in advance if this is a tiresome sort of theme; considering my beginner's lack of competency and self-assuredness, competently-given feedback and assurance from insightful and encouraging users here is more than extremely helpful.
Your proof is "correct", but contains a massive omission. Why should $|\mathcal{P}(X_{N+1})|$ be twice the size of $|\mathcal{P}(X_{N})|$? That's the main part of the proof right there; everything else is the usual induction setup (which you did correctly).
The usual way to complete this gap is to take a set of size $N+1$, and distinguish some arbitrary element; let's paint it blue. Subsets of this set either do or do not contain the blue element; there are $|\mathcal{P}(X_N)|$ of each type, which by induction equals $2^N$. Hence we calculate $2^N+2^N=2^{N+1}$.