The following is taken from Cassels' "Local Fields":
Let $k$ be a field which is complete w.r.t. a non-arch valuation and whose residue field $\rho$ is perfect.
On page 123, in Corollary 3, he makes the following claim:
"the residue field of the algebraic closure of $k\;$ = $\;$the algebraic closure of the residue field of $k$"
He proves this as follows:
1) Take any irreducible polynomial $f$ in $\rho[X].$
2) Lift it to a polynomial $\widetilde{f}$ in $k[X]$.
3) Every root of $\widetilde{f}$ lies in the algebraic closure $\bar{k}$ of $k.$
4) Hence every root of $f$ lies in the residue class field of $\bar{k}.$
5) Done.
Question. Why we are done after step 4?
Denote the residue field of $\bar k$ by $\Omega$. Then $\Omega/ \rho$ is algebraic. Hence, if $\alpha$ is algebraic over $\Omega$, it is algebraic over $\rho$, i.e. a root of some polynomial $f\in \rho[X]$. So steps 1-4 show $\alpha \in \Omega$.