Gromov-Hausdorff distance and the "set of all sets"
I believe the following is correct, but someone may have to come along and correct me. This will further the comment made by Alexander Thumm, above.
There are metric spaces which include isometric copies of every separable metric space. Examples of such spaces are $C([0,1])$, the family of all continuous functions on the unit interval with the sup-norm, and Urysohn's universal metric space $U$. As all compact metric spaces are separable, by fixing such a space $Z$, it should be possible to show that $$d_{GH} (X,Y) = \inf \{ d_H ( i[X], j[Y] ) : i,j\text{ are isometric embeddings of }X,Y\text{, resp., in }Z \}$$
With this in mind, we may consider quotient of the space $\mathcal{K} (Z)$ of all nonempty compact subsets of $Z$ by isometry relation $\sim$. Note that if $K_1,K_2,L_1,L_2$ are compact subsets of $Z$ with $K_1 \sim K_2$ and $L_1 \sim L_2$, then it follows that $d_{GH} ( K_1,L_1 ) = d_{GH} (K_2,L_2)$, and so we may consider $d_{GH}$ as a function from pairs of $\sim$-equivalence classes into the nonnegative reals. The statement that "the set of all compact metric spaces is a metric space under $d_{GH}$" is then an informal interpretation of "$d_{GH}$ is a metric on $\mathcal{K} (Z) / \sim$."
1) How to define rigorously the Gromov-Hausdorff distance between two compact metric spaces.
Let $X$ and $Y$ be two compact metric spaces. As I said in a comment, one does not need to consider all possible isometric embedding of $X$ and $Y$ into any metric space. Any such embedding yields an embedding of $X$ and $Y$ into $X \bigsqcup Y$ (the later being endowed with a pseudo-distance, not a distance).
Hence, one only need to consider all possible pseudo-distances on $X \bigsqcup Y$ whose restriction on $X$ is the distance on $X$, and whose restriction on $Y$ is a distance on $Y$. Then they take the minimum of the Hausdorff distance between $X$ and $Y$ over all such compatible pseudo-distance on $X \bigsqcup Y$. The result is the Gromov-Hausdorff distance.
2) How to work rigorously with a "set of all compact metric spaces".
One proposition says, naively, that "the set of all compact metric spaces, with the Gromov-Hausdorff distance, is a Polish metric space". In particular, it is separable. One can prove the separability by showing that the set of [finite metric spaces with rational distances, up to isometry] is dense in the "set" of [all compact metric spaces, up to isometry].
This proposition, as it stands, lacks rigour. However, we can use it to build a "set of all compact metric spaces, up to isometry", the same way one can construct the real numbers: via completion.
we start from the set $E$ of finite subsets of $\mathbb{N}$, endowed with a metric;
we endow $E$ with the Gromov-Hausdorff pseudo-distance;
we take the equivalent classes of $E$ with respect to the equivalence relation "being isometric". The Gromov-Hausdorff pseudo-distance goes to the quotient, and gives a true distance. This gives a metric space $F$.
we take the completion of $F$ with respect to the Gromov-Hausdorff distance. This gives a Polish metric space, $G$.
for any point $x$ in $G$, we construct a compact metric space $X_x$. This is perhaps the most delicate point; any proof of the "completeness of the set of all compact metric spaces, with the Gromov-Hausdorff distance" should show how to do this.
then we can consider the set of all such compact metric spaces, indexed by $G$. The construction above can be done without any use of the axiom of choice (i.e. it can be done explicitely).
we check that the Gromov-Hausdorff distance between $X_x$ and $X_y$ is the distance between $x$ and $y$ in $G$. Hence we get a set $H$ of compact metric spaces isometric to $G$.
at last, show that any compact metric space is isometric to an element of $H$.
The space $H$ is the one you are looking for (naively, the space of all compact metric spaces, up to isometry).