To elaborate Andreas Thom's comment into an answer:
A note on the space of ultrafilters.
Let $X$ be a topological space and let $pX$ denote its space of ultrafilters, i.e. the Stone-Čech compactification of $X$ with the discrete topology. For any (not necessarily continuous!) map $f : X \to Y$ to a compact space $Y$, there exists a unique continuous map $\tilde f : pX \to Y$, which is given by
$$ \tilde f(p) = \lim_{x \to p} f(x) .$$
Your equivalence relation.
We define two ultrafilters $p,p'\in pX$ to be equivalent if they cannot be distinguished by continuous functions into compact spaces, i.e.
$$ p \sim p' \iff \lim_{x \to p}\ g(x) = \lim_{x \to p'}\ g(x) \quad\text{for all continuous } g:X\to C .$$
We denote the quotient space with $\beta X := pX / \sim$. Thanks to the universality of the space of ultrafilters $pX$, it is clear that every continuous map $g:X \to C$ lifts to a unique, continuous map $\tilde g : \beta X \to C$. What remains to be shown is that $\beta X$ is compact and that the natural map $\iota : X \to \beta X$ is continuous.
$\beta X$ is quasicompact and Hausdorff
As a quotient of a compact space, the space $\beta X$ is quasicompact.
By definition, any two ultrafilters $p \not\sim p'$ can be distinguished by a continuous function $\tilde g : pX \to C$ into a compact space $C$. But $C$ is Hausdorff and we can pull back open neighborhoods from $C$.
The natural map is continuous
This is the hardest part, but it's not too bad. We show that preimages of closed sets are closed.
Let $B\subseteq \beta X$ be a closed set and $A = \iota^{-1}(B)$ be its preimage. Let $\bar A \subseteq X$ be the closure in $X$ and let $y\in \bar A \setminus A$ be a point on the boundary. We have to show that its image $\iota(y)$ is already a member of $B$.
We construct two ultrafilters $p_y$ and $q_y$ as follows:
$p_y := $ the principal ultrafilter on $y$.
$q_y := $ some ultrafilter that contains all open neighborhoods of $y$ and the set $A$. This is possible because every $y$ is from the boundary of $A$, which means that all open neighborhoods of $y$ have nonempty intersection with the set $A$. (Use Zorn's lemma to upgrade the filter generated by these sets to an ultrafilter.)
The key point is that the ultrafilters $p_y$ and $q_y$ converge to the same point $y$ and thus cannot be distinguished by continuous functions from $X$ to some other space. Hence, they are equal in the quotient $\beta X$,
$$ \iota(y) = [p_y] = [q_y] .$$
Using the note above and applying it to $\iota$, we can write this as
$$ \iota(y) = \lim_{x \to p_y} \iota (x) = \lim_{x \to q_y} \iota (x) .$$
But the latter limit must be a member of the set $B$! Otherwise, by the definition of the limit along an ultrafilter, the preimage $A^c =\iota^{-1}(B^c)$ of the open complement $B^c$ would be a member of the ultrafilter $q_y$, which contradicts $A\in q_y$.