Let $X$ be a topological space that enjoys the Hausdorff property, $K \subset X$ be compact, and $\widetilde{X}$ be the Alexandroff compactification of $X$. Is it true that $K \subset \widetilde{X}$ is compact as well?
My attempt to show this:
Let $W_\lambda \subset \widetilde{X}$ be an open cover of $K$, i.e. we have \begin{align} K \subset \bigcup_\lambda W_\lambda. \end{align} Assume that we have $\lambda_0$ s.t. $\infty \in W_{\lambda_0}$, then \begin{align} K \backslash W_{\lambda_0} \subset \bigcup_{\lambda \neq \lambda_0} W_\lambda \subset X. \end{align} Since $W_{\lambda_0}$ is open and contains $\infty$, it is true that there is $C \subset X$ compact, such that $K \backslash W_{\lambda_0} = K \cap C \subset X$ is compact ($X$ is Hausdorff). Hence, there is a finite subcover \begin{align} K \subset \bigcup_{i = 0, \dots, N} W_{\lambda_i}, \end{align} for some natural number $N$ and hence, $K \subset \widetilde{X}$ is compact.
Your proof is okay. But there is nothing to prove.
The Alexandroff compactification $\tilde X$ of any space $X$ has the property that $X$ is an open subspace of $\tilde X$. Thus each $Y \subset X$ receives its original topology when regarded as a subspace of $\tilde X$. But compactness is an intrinsic property of topological spaces and therefore the answer to your question is "yes".