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!

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Your proof is correct in all respect. –  Shailesh Aug 11 '13 at 9:27
Awesome, thanks for the verification. –  cantor_dust Aug 11 '13 at 9:28

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.