[proofplan] We prove directly from the definitions. To show that the subspace $(A,\tau_A)$ is discrete, it suffices to show that every subset of $A$ belongs to $\tau_A$. If $B \subset A$, then $B$ is also a subset of $X$, and discreteness of $X$ makes $B$ open in $X$. Since $B = A \cap B$, the definition of the [subspace topology](/page/Subspace%20Topology) then makes $B$ open in $A$. [/proofplan] [step:Show that every subset of $A$ is open in the subspace topology] Let $B \subset A$ be arbitrary. Since $B \subset A \subset X$, the set $B$ is a subset of $X$. Because $(X,\tau_X)$ is discrete, every subset of $X$ is an element of $\tau_X$, so $B \in \tau_X$. By the definition of the subspace topology on $A$, every set of the form $A \cap U$ with $U \in \tau_X$ belongs to $\tau_A$. Taking $U := B$, we obtain \begin{align*} B = A \cap B, \end{align*} because $B \subset A$. Therefore $B \in \tau_A$. Since $B \subset A$ was arbitrary, every subset of $A$ is open in $\tau_A$. Hence $(A,\tau_A)$ is discrete. [guided] We want to prove that $(A,\tau_A)$ is discrete. By definition, this means that every subset of the underlying set $A$ is open in the topology $\tau_A$. Let $B \subset A$ be an arbitrary subset. The point of choosing an arbitrary $B$ is that proving $B \in \tau_A$ for this one unspecified subset proves the same assertion for all subsets of $A$. Since $A \subset X$, the inclusion $B \subset A$ implies $B \subset X$, so $B$ may also be regarded as a subset of the ambient space $X$. Now we use the hypothesis that $(X,\tau_X)$ is discrete. A [discrete topology](/page/Discrete%20Topology) on $X$ is the topology in which every subset of $X$ is open. Since $B \subset X$, this gives \begin{align*} B \in \tau_X. \end{align*} Next we use the definition of the subspace topology. The topology on $A$ is \begin{align*} \tau_A = \{A \cap U : U \in \tau_X\}. \end{align*} Thus, to prove that $B$ is open in $A$, it is enough to express $B$ as $A \cap U$ for some [open set](/page/Open%20Set) $U \in \tau_X$. The natural choice is $U := B$, which is allowed because we have just shown that $B \in \tau_X$. Since $B \subset A$, intersecting $B$ with $A$ does not change it: \begin{align*} A \cap B = B. \end{align*} Therefore $B = A \cap U$ with $U = B \in \tau_X$, and hence $B \in \tau_A$. Because the subset $B \subset A$ was arbitrary, every subset of $A$ is open in the subspace topology. This is exactly the statement that $(A,\tau_A)$ is discrete. [/guided] [/step]