Here if in the diagram below
Universal property and functoriality
we take a modification that any continuous map $f:X\to K$ where $K$ is compact but not necessarily Hausdorff what would be our $\beta X$ (if $X$ is discrete) in this case instead of all ultrafilters over $X$? Would it always exist?
No, if you drop the Hausdorff condition when talking about the Stone-Cech compactification, then it never exists for any non-compact space. Indeed, suppose $X$ is not compact and suppose there existed an initial continuous map $f:X\to Y$ to a compact space $Y$. Consider the space $K$ obtained by adjoining two points $a,b$ to $X$ and declaring that a set is open in $K$ iff it is either an open subset of $X$ or is equal to all of $K$. Then $K$ is compact and the inclusion map $i:X\to K$ would be continuous, so there would have to be a unique continuous $g:Y\to K$ such that $gf=i$. In particular, this means that the image of $g$ contains all of $X$, and therefore must also contain at least one of $a$ and $b$ since the image of $g$ must be compact and $X$ is not compact. But now define $g':Y\to K$ by $g'(y)=g(y)$ if $g(y)\in X$, $g'(y)=b$ if $g(y)=a$, and $g'(y)=a$ if $g(y)=b$. This $g'$ is still continuous, since the open sets containing $a$ are the same as the open sets containing $b$. Also, for any $x\in X$, $g(f(x))=x\in X$ so $g'(f(x))=x$ as well. That is, $g'f=i$. This contradicts the uniqueness of $g$.
From a categorical perspective, what's going on here is that compact spaces (unlike compact Hausdorff spaces) are not closed under limits in the category of topological spaces, and thus are not a reflective subcategory. The issue is with equalizers: the equalizer of two maps between Hausdorff spaces is closed in the domain, and thus compact if the domain is compact. However, the equalizer of two maps between compact spaces need not be compact.