Prove that if $\mathfrak{J}\subset\mathfrak{J'}$ and $(X,\mathfrak{J'})$ is compact, then $(X,\mathfrak{J})$ is compact. I'm reading Intro to Topology by Mendelson.
The problem statement is in the title.
My attempt at the proof is as follows:
Let $\lbrace U_\alpha\rbrace_{\alpha\in I}$ be an open covering of $X$ such that for each $\alpha\in I$, $U_\alpha\in\mathfrak{J}$. Since $\{U_\alpha\}_{\alpha\in I}$ is also a covering of $X$ in the topology $\mathfrak{J'}$, there then exists a finite subcovering $\{U_{\alpha_i}\}_{i=1}^n$ of $X$, whereby $U_{\alpha_i}\in\mathfrak{J}$ for each $0\leq i\leq n$. Thus, $(X,\mathfrak{J})$ is compact.
My only concern is the last sentence which says all of the subcovering came from the topology $\mathfrak{J}$.
Thanks for any help or feedback!

Your proof is correct.
Sooner or later you will learn that continuous images of compact sets are compact. Then this question can be seen as a special case since $\mathfrak J\subseteq\mathfrak J'$ means that $(X,\mathfrak J)$ is the continuous image of $(X,\mathfrak J')$ under the identity map.
