Relative Algebraic Closure with Algebraically Closed Extension is Algebraic Closure
Theorem
Let $L$ be an algebraically closed field.
Let $L/K$ be a field extension.
Then the relative algebraic closure of $K$ contained in $L$ is an algebraic closure of $K$.
Proof
Let $K'$ denote the relative algebraic closure of $K$ contained in $L$.
By definition, $K' = \{\alpha \in L \mid \alpha \text{ is algebraic over } K\}$.
First we show that $K'$ is a field extension of $K$.
By Field is Algebraic over itself, $K \subseteq K'$.
Since $K$ is a field, it and (therefore) $K'$ are nonzero.
Let $\alpha, \beta \in K'$ be arbitrary.
By definition of generated field extension, $\alpha, \beta \in K(\alpha, \beta) \subseteq L$.
By the field axioms, $\alpha - \beta, \alpha\beta^{-1} \in K(\alpha, \beta) \subseteq L$.
By Finitely Generated Algebraic Extension is Finite and Finite Field Extension is Algebraic, we have that $K(\alpha, \beta)/K$ is algebraic.
Therefore both $\alpha - \beta$ and $\alpha\beta^{-1}$ are algebraic over $K$.
So, by definition, $\alpha - \beta, \alpha\beta^{-1} \in K'$.
Using the Subfield Test/Three-Step, we conclude that $K'$ is a subfield of $L$.
So $K'$ is a field containing $K$, ergo $K'/K$ is a field extension.
Furthermore, $K'$ is algebraically closed.
Indeed, let $p \in K'[X]$, where $K'[X]$ is the ring of polynomials with coefficients in $K'$.
Since $K'[X] \subseteq L[X]$ and, by hypothesis, $L$ is algebraically closed, $p$ has a root in $L$.
Let $\alpha \in L$ be a root of $p$.
Consider the field extension $K'(\alpha)/K'$.
By Finitely Generated Algebraic Extension is Finite and Finite Field Extension is Algebraic, $K'(\alpha)/K'$ is algebraic.
By definition of $K'$, the extension $K'/K$ is algebraic.
Therefore, by Transitivity of Algebraic Extensions, $K'(\alpha)/K$ is algebraic.
Thus $\alpha \in K'(\alpha)$ is algebraic over $K$.
By definition again, we conclude $\alpha \in K'$.
Thus $p$ has a root in $K'$.
We have that $K'/K$ is an algebraically closed algebraic field extension.
Therefore, by definition, $K'$ is an algebraic closure of $K$.
Example
Let $L =$ $\C$ and let $K$ be a subfield of $\C$.
Then the set $\{\alpha \in \C \mid \alpha \text{ is algebraic over } K\}$ is an algebraic closure of $K$.