I’d like to expand on @tomasz”s answer in several ways.
There are really three concepts here: (1) an algebraically closed field; (2) an algebraic closure of a field; and (3) the algebraic closure of one field in another.
I’m going to sweep some delicate points under the rug by choosing a very handy definition of an algebraically closed field: Def. A field $F$ is algebraically closed if it has no proper algebraic extensions. Now, given a field $k$, an algebraic closure of $k$ is an algebraically closed field that is algebraic over $k$. Finally, if $L\supset k$ is an extension of fields, the algebraic closure of $k$ in $L$ is the set of elements of $L$ that are algebraic over $k$. For instance, the algebraic closure of $\mathbb Q$ in $\mathbb R$ is the set (field, really) of all real algebraic numbers. The algebraic closure of $k$ in $L$ is not generally algebraically closed, nor is it an algebraic closure of $k$.
Notice that there is no unique algebraic closure of a field $k$, and that it’s necessary to prove that any two such are isomorphic (as fields containing $k$).
Now to your question. It’s to take $F\subset L$, where $L$ is algebraically closed, and to define $K$ to be the algebraic closure of $F$ in $L$, and show that $K$ is also algebraically closed. Well: $K$ is certainly an algebraic extension of $F$ (all elements are algebraic over $F$). So, let $K'$ be an algebraic extension of $K$. We have $F\subset K\subset K'$, both inclusions being algebraic. But algebraic over algebraic is still algebraic, so $K'$ is algebraic over $F$, i.e. every element of $K'$ is algebraic over $F$, so every element of $K'$ is in $K$, and you’ve shown that $K$ has no proper algebraic extensions.
You are too generous with equality and you start by assuming a not yet specifically defined $\overline K$.
I'd suggest you start with
Lemma. Let $K$ be a field, $L/K$ algebraic (but not necessarily algebraically closed), $M/K$ algebraically closed (but not necessarily algebraic). Then there exists at least one field homomorphism $f\colon L\to M$.
Proof sketch. Consider the set of all field homomorphisms $\phi\colon L'\to M$ where $K\subseteq L'\subseteq L$ and define a partial order on this set by saying $\phi\le \psi$ if $\psi$ is an extension of $\phi$ (i.e., the domain of $\phi$ is a subfield of the domain of $\psi$ and $\phi$ is the restriction of $\psi$ to that subfield). Verify that Zorn's lemma can be applied. Pick a maximal element $\phi_\max\colon L_\max\to M$ and show that in fact $L_\max=L$ (using that $L$ is algebraic and $M$ is algebraically closed).
Once you have that, apply the lemma to the case that $L$ and $M$ are both algebraic and algebraically closed and show that any $f$ the lemma gives you is in fact onto.
Best Answer
The algebraic closure $\mathbb A$ of $\mathbb Q$ is the field of algebraic numbers, which consists of those complex numbers which are roots of some non-zero polynomial in one variable with rational coefficients. It is a countable set and therefore $\mathbb{A}\varsubsetneq\mathbb C$.