Give topological space $(X,\tau)$, $E \subset X$, $\{x_{\alpha}\} \subset E, x\in E$ and $(E,\tau_{E})$ is subset topological space of $(X,\tau)$.
I have proved that: If $x_{\alpha} \xrightarrow{(E,\tau_{E})} x$ then $x_{\alpha} \xrightarrow{(X,\tau)} x$.
(Here is my proof)
Give $V$ is neighborhood of $x$ in $(X,\tau)$.
There exist $G \in \tau$ that $x \in G \subset V$.
Therefore, $x \in G \cap E \subset V \cap E$.
So, $V \cap E$ is a neighborhood of $x$ in $(E,\tau_{E})$.
Since $x_{\alpha} \xrightarrow{(E,\tau_{E})} x$ then exist $\alpha_{o}$ that if $\forall \alpha \ge \alpha_{o} \Rightarrow x_{\alpha} \in V \cap E \subset V$
Therefore, $x_{\alpha} \xrightarrow{(X,\tau)} x$.
I wonder that the converse is correct or not?
The forward proof is correct, though somewhat verbose.
You can also just note that $i(x)=x$ as a map from $(E,\tau_E) \to (X, \tau)$ is continuous by definition of the subspace topology, and so preserves net convergence; i.e. the same net in the large space has the same limit.
The converse is similar: if $x_\alpha \to x$ in $(X,\tau)$ with all $x_\alpha$ and $x$ in $E$, let $O \ni x$ be open in $E$, so $O = O' \cap E$ with $O'$ open in $X$ and we note $x \in O'$ so some tail of the net sits inside $O'$, and so inside $O=O'\cap E$ as well. Hence $x_\alpha \to x$ in $(E,\tau_E)$ too.