I am unsure if this is correct. Any help/ guidance is appreciated!
We first verify that the inequality holds for the initial values of the sequence, which are $a_0 = a_1 = a_2 = 2$
$$P(0) = 2 < 4$$ $$P(1) = 2 < 8$$ $$P(2) = 2 < 16$$
Show that for every integer $k \geq 2$, if $P(i)$ is true for each integer $i$ from $0$ through $k$, then $P(k+1)$ is true:
Let $k$ be any integer with $k \geq 2$, and suppose that $a_i \leq i$ for each integer $i$ with $0 \leq i \leq k$.
We must show that
$$a_{k+1} < 2^{(k+1) + 2}$$
Now
$$a_{k+1} = a_k + a_{k-1} + a_{k-2}$$ $$= 2^{k+2} + 2^{k+1} + 2^{k}$$ $$= 2^k * 2^2 + 2^k * 2 + 2^k$$ $$= 2^k(2^2 * 2)$$ $$= 2^{k+3}$$
Thus $a_{k+1} < 2^{(k+1) + 2}$
The first version of your proof was nearly correct, but the current version is slightly flawed. Here is the corrected version. Let $a_n = a_{n - 1} + a_{n - 2} + a_{n - 3}$ and $a_0,a_1,a_2 = 2$. We prove that $a_n < 2^{n + 2}$, $\forall n \geq 0$. The inequality holds for the three base cases:
$$a_0 = 2 < 2^2 = 4$$ $$a_1 = 2 < 2^3 = 8$$ $$a_2 = 2 < 2^4 = 16$$
Now we consider $n \geq 2$:
$$a_{n + 1} = a_n + a_{n - 1} + a_{n - 2}$$ $$ < 2^{n + 2} + 2^{n + 1} + 2^{n}$$ $$ = 2^n \left( 2^2 + 2 + 1 \right) = 7\cdot2^n$$ $$< 8 \cdot 2^n = 2^n \left( 2^3 \right) = 2^{n + 3}$$
And so we conclude that $a_{n + 1} < 2^{n + 3}$ and since the three base cases hold it follows from strong induction that $a_n < 2^{n + 2}$, $\forall n \geq 0$.