The fundamental example of algebraically closed field is $\C$. If $\Omega$ is algebraically closed containing $k$, then the set of elements of $\Omega$ algebraic over $k$ is a field, and this set is also an algebraic closure of $k$.
Example: Let $A=\{\alpha\in\C\text{ such that }\alpha$ algebraic over $\Q\}$, also known as the set of algebraic numbers in $\C$, then $A$ is a subfield of $\C$ and is algebraic closure of $\Q$, $A$ is denoted as $\wbr{\Q}$.

Proof Let $f\in\wbr{k}[X]$ be a non constant polynomial. It suffices to show that there is a root in $\wbr{k}.$ The field $L$ generated by the coefficients of $f$ is of finite dimension on $k$, since the coefficients of $f$ are algebraic over $k$. Thus, the $k$ algebra $A=L[X]/\lr{f}$ is of finite dimension over $k$, where the dimension is $\deg f\cdot\dim_kL$ (via tower law). Consider the evaluation map $k[T]\ra A$, the morphism is never injective since $\dim_k(k[T])=\infty$. Thus there is a non zero $g\in k[T]$ which will get mapped to zero, or in other words $f|g$, in particular $g$ is non constant (since $\deg f>0$). But, $g$ has coefficients in $k$ and thus will split in algebraic closure $\wbr{k}$. Thus, it is the same for $f$ which divides $g$.

Note that the isomorphism whose existence is given by the theorem is far from unique as we will see: we can even prove that an algebraically closed field admits a infinity of automorphisms. We will first prove the existence, then the uniqueness which follows from fundamental theorem of extension of morphisms.

Let us start by building a huge algebra $A$ in which every polynomial has a root. Let $c(P)$ denote the dominant coefficient of every non zero polynomial. Start by considering the algebra of polynomials with large number of indeterminates (indeterminate for each possible root of $P$) $$ A=k[X_{P,i}] \text{ where }1\leq i\leq \deg P$$ Let $\gamma(i,P)$ denote the coefficients of the following polynomial in $X$, $$P(X)-c(P)\prod_{i=1}^{\deg P}(X-X_{P,i}) $$ Let $I$ be the ideal generated by $\gamma(i,P)$ with $P$ non constant. In case $P$ is a non zero constant, we have $\gamma(0,P)=0$. Ideal $I$ is proper.

Let $J$ be a maximal ideal of $A$ containing $I$ (use Zorn's Lemma for construction) and $L$ be the field $A/J$. By construction, all non constant polynomials $P\in k[X]$ split in $L$, the roots being the images of $X_{P,i}$. All these roots are algebraic on $k$. But they generate $L$ as $k$-algebra. So we have $L$ algebraic on $k$.

Proof Let $E$ be a non empty subset of couple $(L,\sigma)$ where $L$ is a subfield of $K$ containing $k$ and $\sigma$ is an embedding $\sigma:L\ra \Omega$ making $\Omega$ an $L$ algebra. The extension of the embedding defines an order on $E$ that is clearly an inductive set. Let then $(L,\sigma)$ be a maximal element.
We now show $L=K$. Let $x\in K$, as $x$ is algebraic over $k$ it is also on $L$. Let $P(X)=\sum_ia_iX^i$ be the minimal polynomial of $x$ on $L$ so that the evaluation at $x$ is identfied with $L[X]/\lr{P}=L[x]$. Let $y$ be a root of $P^\sigma(X)=\sum\sigma(a_i)X^i$ in $\Omega$. There exists a unique $L$ morphism $L[X]/\lr{P}\ra\Omega$ that sends $X$ to $y$ because the image of $P$ in $\Omega$ is by definition $P^\sigma(y)=0$, where the $k$ morphism $L[x]\ra\Omega$ extends $\sigma$. By, maximality of $L$, we deduce $x\in L$.

Remark: $σ$ allows us to identify $L$ with $σ (L)$. From now on, we will not distinguish between $L$ and $σ (L)$. Note also that if $K$ is a finite extension, then the notion of dimension makes it possible to show the existence of $L$ without invoking Zorn's lemma.

Proof Consider $K_1$ as algebraic and $K_2$ as algebraically closed, the theorem above assures the existence of an embedding of $K_1$ in $K_2$. With the preceding notations, the choice of an embedding of $K_1$ in $K_2$ permits to see $K_1$ as a subfield of $K_2$. Now consider $(k,K,\Omega)$ and $(K_1,K_2,K_1)$, we deduce the existence of $\tau\in\Hom_{K_1}(K_2,K_1)$, in other words the following diagram is commutes. As $\tau $ is a morphism of fields, it is injective. The equality $\tau\circ\sigma=$Id assures the surjectivity. $\tau$ and $\sigma$ are inverses of each other.

Proof If $y\in\Omega$ is a conjugate of $x$, there exists $\sigma\in \Hom_k(k[x],\Omega)$ such that $\sigma(x)=y$ (this follows from the lemma which gives the correspondence $\Root_f\Omega\leftrightarrow \Hom_k(k_f,\Omega) $, in our case $k_f=k[X]/P=k[x]$ where $x$ is root of the polynomial $P$). Now we just need to extend $\sigma$ to all of $K$, which is possible by the Theorem. Inversely, if $\sigma\in\Hom_k(K,\Omega)$, it leaves minimal polynomial of $x$ on $k$ invariant. Thus, it only permutes the roots, which are the conjugates of $x$ by definition.

Let $K$ be a sub field of $\Omega$ algebraically closed. Let $P$ be a polynomial of $k[X]$ not necessarily irreducible. The field of roots of $P$ is a subfield of $\Omega$ generated by the roots of $P$ in $\Omega$. This is the smallest subfield of $\Omega$ in which $P$ splits. Of course, it is contained in the algebraic closure of $k$ in $\Omega$ and does not depend upon $\Omega$.
  The philosophy is to fix an algebraic closure in which we work, this then allows us to talk about the field of roots.