Given set $S=\{\alpha,\beta,\gamma\}^\Bbb{N}$ of sequences of $3$ symbols and function $d : S \times S \to \Bbb{R}$ $$d(a,b)=\begin{cases}\frac3{i(a,b)} & a\neq b\\ 0 & a=b\end{cases}$$ where $i(a,b)=\min\{k \in \Bbb{N} \mid a_k\neq b_k\}$.
a) Check if $A=\{a\in S \mid a_7 = \gamma\}$ is bounded,closed and compact.
b) Check if the metric space $(S,d)$ is connected.
Now for a) we prove that $A$ is closed by proving that $A^c=\{a\in S \mid a_7\neq \gamma\}$ is open we have that if $a\in A^c$ and $b\in A$ then $d(a,b)\geq \frac37$ so for every $a\in A^c$ we have that $B(a,3/7)\subseteq A^c$ hence $A^c$ is open.
$d(a,b)\leq 3$ hence $A\subseteq B(a,4)$ hence $A$ is bounded. I don't know how to prove that the set is compact.
For b) I either have to find two open sets $A_1,A_2$ such that $A_1\cup A_2=A,A_1\cap A_2=\emptyset$ or prove such sets don't exist or to find/prove it doesn't exist a continuous surjective function such that $f : S \to \{0,1\}$ with the discrete metric.
Proof of compactness of $A$: $A$ is a closed subset of $S$ and $S$ itself is compact: Consider $S$ with the product topology induced by the discrete topology on $\{\alpha, \beta, \gamma\}$. Without loss of generality take $\alpha, \beta, \gamma $ to be real numbers, say $1,2$ and $3$. This topology is induced by the metric $D((a_n),(b_n))=\sum \frac {|a_n-b_n|}{2^{n} (1+|a_n-b_n|)}$. I leave it to you to verify that the identity map $f$ from $(S,D)$ to $(S,d)$ is continuous. [$D((a_n),(b_n))< \frac 1 {2^{n+1}}$ implies that $a_i=b_i$ for $1 \leq i \leq n$ which implies $d((a_n),(b_n))<\frac 3 n$.] Being a continuous image of a compact space the given space $(S,d)$ is compact.
$S$ is not connected: the map $f(a_n)=a_1$ is continuous from $S$ onto $\{\alpha, \beta,\gamma\}$ with the discrete metric and its image is not connected. Continuous image of a connected space is connected, so $S$ is not connected. [Continuity follows from the fact that $f(a_1)=f(b_1)$ whenever $d((a_n)(b_n)) <\frac 3 2]$.