Definition of compact in this context:
A set $A$ is said to be compact if for any sequence in $A$ there is a convergent subsequence which converges to a value which belongs to $A$.
Prove that if $F$ is a countable collection of open sets that cover a compact set $A$ then a finite number of open sets in $F$ cover $A$. $$$$Suppose that $$F=\bigcup_{k=1}^{\infty}I_k$$ be the countable collection of open sets $I_k$ that cover $A$, I will prove that for some $n$ the collection $$\bigcup_{k=1}^{n}I_k$$ covers $A$. Suppose that this property does not hold then for every $n$ there exists a $a_n$ member of $A$ such that $a_n$ does not belong to the collection $$\bigcup_{k=1}^{n}I_k$$. Now as $a_n$ is a member of set $A$ and as $A$ is compact and hence bounded, so $a_n$ is also bounded, so there exists a subsequence say $b_k$ of $a_n$ such that $b_k$ converges to $b$ which also belong to set $A$ as the set $A$ is compact. Now for every $M$ if we chose any $n>M$ then we see that $a_n$ doesn't belong to the collection $$\bigcup_{k=1}^nI_k$$ and as $n>M$ so $a_n$ also doesn't belong to the collection $$\bigcup_{k=1}^MI_k$$. Now for every $M$ if we chose any $k_0>M$ then as $b_k$ is the subsequence of $a_n$ so there exists a $n_0>k_0$ such that $$a_{n_0}=b_{k_0}$$ and also as $n_0>k_0>M$ and hence as shown above $a_{n_0}=b_{k_0}$ doesn't belong to the collection $$\bigcup_{k=1}^{M}I_k.$$ Now as $b$ also belongs to $A$ so there is some $p_0$ such that $b$ lies in the open set $I_{p_0}$. Now as $I_{p_0}$ is an open set so there exists a $\epsilon>0$ such that the interval $$(b-\epsilon, b+\epsilon)$$ lies in the set $I_{p_0}$. Now as $b_k$ converges to $b$, so there exists a $K_1$ such that for every $k>K_1$ we have $$b-\epsilon < b_k < b+\epsilon$$. Also for $M=p_0$ for every $k>M=p_0$ as shown above $b_k$ doesn't belong to the collection $$\bigcup_{k=1}^{p_0}I_k$$. Now chose $$p=\text{max}(p_0, K_1),$$ then for every $k>p$ we get that $$b-\epsilon < b_k < b+\epsilon$$ and also $b_k$ doesn't belong to the collection $$\bigcup_{k=1}^{p}I_k,$$ but the interval $(b-\epsilon, b+\epsilon)$ lies in the open set $I_{p_0}$ which belongs to the collection $$\bigcup_{k=1}^{p}I_k$$ and hence a contradiction.
Is my proof correct?
I suspect your answer is okay, but quite cumbersome.
Check yourself on my (more concise) answer.
Assume that $A\subseteq\bigcup_{k=1}^{\infty}I_k$ where the $I_k$ are open sets and assume that for every $n$ we have: $$A-\bigcup_{k=1}^{n}I_{k}\neq\varnothing$$
For every $n$ let $a_{n}\in A-\bigcup_{k=1}^{n}I_{k}$.
Then let $n_{1}<n_{2}<\dots$ such that $\left(a_{n_{i}}\right)_{i}$ is a convergent sequence with limit $a:=\lim_{i\to\infty}a_{n_i}\in A$.
Then $a\in I_{m}$ for some $m$ and - because $I_{m}$ is open - consequently some $i_{0}$ exists with $i\geq i_{0}\implies a_{n_{i}}\in I_{m}$.
This however demands $n_{i}<m$ for every $i$ and a contradiction is found.
We conclude that our assumption was wrong which means that $A\subseteq\bigcup_{k=1}^nI_k$ for some $n$.