Showing a set is compact I need help with the question 7/b part in the image below. About the point 0, any open set contains points of the other non zero set , except for finitely many points. But then, I am facing difficulty in adjoining the rest of the open covers and their form.

 A: You can show sequencial compactness which in a metric space like $\mathbb{R}$ is equivalent to compactness. If you have $s_i=\frac{1}{m_i}+\frac{1}{n_i}$ then there are 3 cases:
1) assume $m_i,n_i$ are bounded and by the pigeonhole principle you can prove that there exists a $k$ such that $s_k$ appears infinitelly many times. Pick this subsequence then.
2) assume WLOG $m_i$ is bounded and $n_i$ is not. Take an infinite increasing subsequence of $n_i$ and use pigeonhole principle again to conclude the existence of infinite values $s_i=\frac{1}{m_k}+\frac{1}{n_i}$ where $m_k$ is constant and $n_i$ is increasing. This subsequence converges to $\frac{1}{m_k}$
3) assume $m_i,n_i$ are both unbounded. Take any infinite increasing sequence $n_i$ and then either the corresponding $m_i$ form an unbounded sequence from which you can then pick an increasing $m_j$ subsequence and take $s_j=\frac{1}{m_j}+\frac{1}{n_j}$ with $\lim_{j\to\infty}n_j=0$ or the corresponding values $m_i$ form a finite set and you can procceed as in 2).
A: Let $\mathcal{U}$ be an open cover of $S$ by open sets of the reals. Let $U_0 \in \mathcal{U}$ be such that $0 \in U_0$ and find $r>0$ such that $(-r,r) \subseteq U_0$. Then, setting $M= \frac{2}{r}$ we find that all points $x(n,m) = \frac{1}{n}+\frac{1}{m}$ with $n,m > M$ already obey $x < r$ and so $x \in U_0$, so we only need to take care of the points $x(n,m)$ with $n \le M$ or $m \le M$ (by symmetry we only need to consider the $n \le M$ case, as $x(n,m)=x(m,n)$). 
Now, for $n \le M$ we note that $\frac{1}{n} = \frac{1}{2n}+\frac{1}{2n} \in S$ and so $\frac{1}{n}$ must be covered by some $U_n \in \mathcal{U}$ as well. As $\frac{1}{n}+\frac{1}{m} \to \frac{1}{n}$ as $m \to \infty$, all but finitely many $x(n,m)$, $x(n,m) \in U_n$. 
So the final finite subcover consists of $U_0$, the finitely many $U_n$ for $n \le M$ and some finitely many $U_x$ for those $x$ that are the (finite times finitely many) not-yet-covered points for the aforementioned finitely many ones.
A: Consider subset $P = \{(\frac{1}{m}, \frac{1}{n} | n, m \in \mathbb N\} \cup \{(0, 0)\}$ of plane. $P$ is compact: any convergent sequence convergence in both coordinates, and convergent sequence with elements $\{\frac{1}{n}\} \cup \{0\}$ is either eventually constant, or converges to $0$, so it converges to element from $\{\frac{1}{n}\}\cup \{0\}$.
Now, let us take function $f: \mathbb{R}^2 \to \mathbb{R}$, $f(x, y) = x + y$. Then $f$ is continuous and $S = f(P)$. And continuous image of a compact is a compact.
