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Consider $\mathcal{A}=\left\{1,2,\ldots,k\right\}$ with discrete topology and $X=\mathcal{A}^{\mathbb{Z}}$ with associated product topology. So called cylinder sets $[a_m,\ldots,a_n]:=\left\{x\in X: x_j=a_j\text{ for }m\leqslant j\leqslant n\right\}$ form a base of the product topology.

Am I right that each subset $S\subset X$ is compact?

My argument is that $X$ is compact by Tychonoff's Theorem and the complement of each subset $S\subset X$ is open, since for $x\in X\setminus S$, we always have an open set containing $x$: Simply choose some cylinder set containing $x$, e.g. $[x_{-1},x_0,x_1]$.

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    This is not correct. Consider instead the homeomorphic $\mathcal{A}^{\mathbb{N}}$. This maps continuously onto $[0, 1]$ by $(a_n)_{n \in \mathbb{N}} \mapsto \sum_{n = 1}^{\infty} (a_n - 1) k^{-n}$ (i.e. by considering a sequence as a $k$-ary expansion). If every subset of $X$ was compact, then so would be every subset of $[0, 1]$. – 2017-02-11
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    Another way to see it would be that $X$ is Hausdorff, and every compact subset of a Hausdorff space is closed. Thus if a Hausdorff space has only compact subsets, it must be discrete. – 2017-02-11
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    Ok, for example, $Y=\bigcap_{n\geq 1}T^n(X)$ for some continuous map $T\colon X\to X$ is compact. Andnow I need to show that some special subset of $Y$ is also compact. – 2017-02-11
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    I'm not sure I understand your comment. But I consolidated this all into an answer. – 2017-02-11
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    The origin of my question is that I do not know how to prove that special subsets of $Y$ are compact or not. – 2017-02-11
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    Look at my answer. We say a set is compact if for *every* open cover, *there exists* a finite sub-cover. Very often when we want to show some property holds, it will be easier to prove than disprove (or vice-versus). For compactness, it's often the latter. The negation of compactness is *there exists* an open cover such that *there does not exist* a finite sub-cover. So showing non-compactness is usually easier, as it means we only have to show an example of an open cover with no finite sub-cover, as I did in my answer. – 2017-02-11
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    It's also worth noting that proving compactness is often a very non-constructive endeavor. The general formula for proving a set is compact is more properly to show it's not not compact. More specifically, it involves showing the impossibility of showing the set in question is not compact. A good elementary example of this is the Heine-Borel theorem. In its simplest form, the theorem states that a subset of $\mathbb{R}$ is compact if and only if it's closed and bounded, but looking at the actual proof method is instructive for how we sometimes show compactness. – 2017-02-12

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This is not correct. First, we can see at least two ways a priori that something's amiss.

First: Consider instead the homeomorphic $\mathcal{A}^{\mathbb{N}}$. This maps continuously onto $[0, 1]$ by $(a_n)_{n \in \mathbb{N}} \mapsto \sum_{n = 1}^{\infty} (a_n - 1) k^{-n}$ (i.e. by considering a sequence as a $k$-ary expansion). If every subset of $X$ was compact, then so would be every subset of $[0, 1]$.

Second: Another way to see it would be that $X$ is Hausdorff, and every compact subset of a Hausdorff space is closed. Thus if a Hausdorff space has only compact subsets, it must be discrete.

These are all indirect ways about it, but to construct a counterexample we could just take $X \setminus \{ \mathbf{x} \}$ for some element $\mathbf{x}$. I'm gonna proceed as if this were $\mathcal{A}^{\mathbb{N}}$ for convenience. Set $\mathbf{x} = (x_n)_{n \in \mathbb{N}}$, and let \begin{align*} U_m & = \bigcup_{(y_1, \ldots, y_m) \neq (x_1, \ldots, x_m)} [y_1 , \ldots , y_m] . \end{align*} Then we can see that $\{ U_m : m \in \mathbb{N} \}$ is an open cover of $X \setminus \{ \mathbf{x} \}$, but if we look at, say, only $U_1 \cup \cdots \cup U_M$, then we can look at $\mathbf{z} = (z_n)_{n \in \mathbb{N}}$, where $$z_n = \begin{cases} x_n & n \neq M + 1 \\ c & n = M + 1 , \end{cases}$$ and $c \neq x_{M + 1}$. Then $\mathbf{z} \in X \setminus \{ \mathbf{x} \}$, but $\mathbf{z} \not \in U_1 \cup \cdots \cup U_M$. Thus $\{ U_m : m \in \mathbb{N} \}$ is an open cover of $X \setminus \{ \mathbf{x} \}$ which admits no finite subcover.

While you're at it, you can see $X \setminus \{ \mathbf{x} \}$ isn't closed, as any cylinder about $\mathbf{x}$ would contain other (in fact uncountably many) elements.