Neighborhood in Topological Subspace

Theorem
Let $\left({X, \tau}\right)$ be a topological space.

Let $S \subseteq X$ be a subset of $X$ and let $\tau_S$ denote the subspace topology on it.

Let $x \in X$ be an arbitrary point of $X$.

Let $E \subseteq S$.

Then $E$ is a neighborhood of $x$ in $\left({S, \tau_S}\right) \iff \exists D \subseteq X$ such that $E$ is a neighborhood of $x$ in $X$ and $E = D \cap S$.

Proof
$\left(\implies\right)$ Suppose that $E$ is a neighborhood of $x$ in $\left({S, \tau_S}\right)$.

By the definition of neighborhood, $\exists U \in \tau_S : x \in U \subseteq E$.

Now, by the definition of the subspace topology, $\exists V \in \tau : U = V \cap S$. Take $D := V \cup E$.

Then $D$ clearly is a neighborhood of $x$ in $X$, because $V \subseteq D$ and $V \in \tau$.

Therefore it holds true that $D \cap S = \left(V \cap S\right) \cup \left(E \cap S\right) = U \cup E = E$.

$\left(\impliedby\right)$ Suppose that $\exists D \subseteq X$ such that E is a neighborhood of $x$ in $\left({X, \tau}\right)$ and $E = D \cap S$.

Then, by the definition of neighborhood, $\exists V \in \tau : x \in V \subseteq D$.

As a consequence, $E = D \cap S \supseteq V \cap S \in \tau_S$, that is, $E$ is indeed a neighborhood of $x$ in $\left({S, \tau_S}\right)$.