As you mentioned, a continuous map $f:X\to Y$ doesn't induce a continuous map $\alpha X\to\alpha Y$ unless it is proper: if you want $\alpha f$ to act as
$$
\alpha f(x) = \begin{cases} f(x), & x\in X \\ \infty, & x=\infty \end{cases}
$$
then this is only continuous if it sends open neighbourhoods of $\infty\in\alpha Y$ to open neighbourhoods of $\infty\in\alpha X$. Since these neighbourhoods are complements of compact sets, this means you need $f$ to be proper.
Therefore, $\alpha$ defines a functor $\def\Top{\mathbf{Top}}\Top_{\mathrm{prop}}\to\Top$.
However, note that $\alpha X$ need not be Hausdorff in general, even if $X$ is Hausdorff! In fact, $\alpha X$ is Hausdorff if and only if $X$ is Hausdorff and locally compact (local compactness ensures that $\infty$ can be distinguished from the points of $X$). Therefore, if you want $\alpha$ to map into compact Hausdorff spaces, you need to restrict your domain, in which case you get a functor $\alpha:\mathbf{LCHaus}_{\mathrm{prop}}\to\def\cHaus{\mathbf{cHaus}}\cHaus$.
Let's now address your questions.
I'll start with (3). In more elementary terms, $\cHaus$ being a (full) subcategory of $\Top_{\mathrm{prop}}$ amounts to checking that continuous maps between closed Hausdorff spaces are proper, and indeed the answer is yes:
Claim. If $X$ is compact and $Y$ is Hausdorff, then any continuous function $f:X\to Y$ is proper.
Proof. If $K\subseteq Y$ is compact, then it is in particular closed since $Y$ is Hausdorff. Therefore, $f^{-1}(K)$ is closed because $f$ is continuous, but closed subsets of a compact space are compact. $\blacksquare$
Now, question (1) merges with (2), since $\beta$ can only satisfy the same universal property if it is left adjoint to the forgetful functor $i:\cHaus\to\Top_{\mathrm{prop}}$.
For this to hold, we need a natural correspondence
$$
\def\Hom{\operatorname{Hom}}\Hom_{\mathrm{prop}}(X,i(Y)) \cong \Hom(\beta X,Y)
$$
for $Y$ compact Hausdorff, and $X$ arbitrary.
However, by the ordinary universal property of $\beta$, we have a natural correspondence $\Hom(\beta X,Y)\cong\Hom(X,Y)$ (where the right hand side consists of all continuous functions $X\to Y$).
Therefore, $\beta X$ satisfies the same universal property in $\Top_{\mathrm{prop}}$ if and only if every continuous function $X\to Y$ into a compact Hausdorff space $Y$ is proper.
By the earlier claim, $\beta X$ thus satisfies the same universal property if $X$ is already compact to begin with.
Conversely, if $f:X\to Y$ is a proper continuous function with $Y$ compact, then $f^{-1}(Y)=X$ must be compact also.
Conclusion. $\beta X$ satisfies the same universal property in $\Top_{\mathrm{prop}}$ if and only if $X$ is compact.
Let me be clear what the conclusion is saying: if $X\to Y$ is a proper continuous function into a compact Hausdorff space $Y$, then certainly it factors uniquely through the compactification $X\to\beta X\to Y$ (this is the ordinary universal property), and moreover $\beta X\to Y$ is certainly proper as well.
Ultimately, the problem is that $X\to\beta X$ is not proper in general, and the conclusion is really saying that $X\to \beta X$ is proper if and only if $X$ is compact.
