In my lecture notes there is written an equation
$$\mathbb{E}[X_{t+1} \mid X_t =x ] \leq (1-\delta) x,$$
(how this equation is derived, does not really matter here).
Then the next part is:
"Taking the expectation on both sides,
$$\mathbb{E}[\mathbb{E}[X_{t+1} \mid X_t ] \mid X_0 = s_0] \leq (1-\delta) \mathbb{E}[X_t \mid X_0 = s_0]. \quad (1)$$
By the definition of a conditional expectation, we have
$$\mathbb{E}[X_{t+1} \mid X_0 = s_0]= \mathbb{E}[\mathbb{E}[X_{t+1} \mid X_t ] \mid X_0 = s_0], \quad (2)$$
so
$$ \mathbb{E}[X_{t+1} \mid X_0=s_0] \leq (1-\delta) \mathbb{E}[X_t \mid X_0 = s_0]." \quad (3)$$
So I am really confused about this derivation and need some help further explaining these steps. The step from (2) to (3), however, is obvious, since it is only putting (1) into (2). I am a beginner in working with conditional expectations, so this is probably why I have no idea how to tackle such a problem. Thank you very much for your help.
Edit: Maybe one sentence about what confuses me the most: Taking the conditional expectation (conditioned on $X_0=s_0$) seems to be legit, but why does the conditioning on $X_t=x$ (in the first equation) disappear?
That first (un-numbered) inequality is valid for every possible value $x$ of $X_t$. If so, then the inequality is valid for $X_t$ itself: $$ E(X_{t+1}|X_t) \le (1-\delta)X_t\tag0$$ That's how the conditioning on $X_t=x$ disappeared; it was replaced by the above statement about $X_t$, omitted from your lecture notes, which I've labeled ($0$).
Line ($1$) is a straightforward application of $E(\cdots|X_0=s_0)$ to both sides of ($0$). Line ($2$) follows from the def of conditional expectation because $\sigma(X_0)$ is a coarser $\sigma$-algebra than $\sigma(X_t)$ (the "tower" property of conditional expectation).