$|\cdot|_p$ denotes the p-adic valuation throughout.
Krasner's Lemma: Let $a, b \in \overline{\mathbb{Q}}_p = $ algebraic closure of $ \mathbb{Q}_p$ with $|b-a|_p < |a_i - a|_p$ where the $a_i \neq a$ are the conjugates of a. Then $\mathbb{Q}_p(a) \subset \mathbb{Q}_p(b) $
Proof: Let $K = \mathbb{Q}_p(b)$, and suppose $a \notin K.$ Then $\exists a_i \notin K, a_i \neq a$ and $a_i$ a conjugate of $a$ over $K$. There exists an isomorphism $\sigma: \mathbb{Q}_p(a) \to \mathbb{Q}_p(a_i)$ that fixes K with $\sigma(a) = a_i$. We have $|b-a_i|_p = |b-a|_p$ because of uniqueness of norms hence
$|a_i - a|_p \leq max(|a_i-b|_p, |b-a|_p)=|b-a|_p < |a_i-a|_p$ a contradiction as required.
This proof makes sense to me but I think it only proves that $\mathbb{Q}_p(a) \subseteq \mathbb{Q}_p(b)$ i.e. the statement of the lemma without the strict inequality. Indeed versions of Krasner's lemma in more generality than this do not include the strict inequality however I have seen the version given here in a few sources which suggest it's not a mistake. How do you deduce that $\mathbb{Q}_p(a) \neq \mathbb{Q}_p(b)$?
The proof given here is taken from p.69 of Koblitz p-adic numbers, p-adic analysis and zeta-functions which can be found here: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.461.4588&rep=rep1&type=pdf
I think Koblitz's notation $ X \subset Y$ means, $X$ is contained in or equal to $Y$, see https://en.wikipedia.org/wiki/Subset, the section on "⊂ and ⊃ symbols".
Certainly it should be impossible to prove a strict containment from these assumptions: If $a = b$ then $|a-b|_p = 0$ so the hypotheses are satisfied and $\mathbf Q_p(a) = \mathbf Q_p(b)$.