Let $x_1=1$, $x_2=2$, and $x_n=\frac{1}{2}(x_{n-1}+x_{n-2})$. Show using strong induction that $x_n\in [1,2]$ for all natural $n$.
So I know just from inequalities that if $a<b$ then $a<\frac{a+b}{2}<b$ so intuitively it makes sense since we are having a sequence a numbers whose $n$-th entry is the average of the $(n-1)$-th and the $(n-2)$-th term. I can also show using recursive properties that a formula in $n$ exists and can bound that. But I want to use "strong induction". So we just have to assume that (as long as the base case holds) the assumption above holds for all $k\le n$ and then check the $(k+1)$-th term meets the conditions of the assumption.
So for $x_3$, we have
$$x_3=\frac{1}{2}(x_2+x_1)=\frac{1}{2}(3)=\frac{3}{2}\in[1,2]$$
Assuming now that for $k\le n$, we have that $x_k\in [1,2]$. For $x_{k+1}$,
$$x_{k+1}=\frac{1}{2}(x_k+x_{k-1}) \qquad \Rightarrow \qquad 2x_{k+1}=x_k+x_{k-1}$$
But now, from our assumption, it must be for all $x_k$ where $k\le n$, that $x_k+x_{k-1}\in[2,4]$ Thus
$$2\le x_k+x_{k-1}\le4 \qquad \Rightarrow \qquad 2\le 2x_{k+1}\le4 \qquad \Rightarrow \qquad 1\le x_{k+1}\le 2$$
Thus, $x_n\in[1,2]$ for all $n$. Is this proof sufficient?
Note that for base case, actually you just need to check for $x_1$ and $x_2$. There is no need to check for $x_3$ explicitly.
Your induction step is fine.
In general, if $x_1 = c$ and $x_2=d$, $c \le d$ and $x_{n} = \lambda x_{n-1} + (1-\lambda)x_{n-2} $ where $\lambda \in [0,1]$, then we have $c \le x_n \le d $ and we can see this from strong induction, where we can use your idea of proof.
The statement is true for $x_1$ and $x_2$.
If $c \le x_{k-1} \le d$ and $c \le x_{k} \le d$, then we have
$\lambda c \le \lambda x_{k-1} \le \lambda d$ and $(1-\lambda)c \le (1-\lambda)x_{k} \le (1-\lambda)d$, then by adding them up, we have
$$c \le x_{k+1} \le d.$$