Let $X$ and $Y$ be two Banach spaces and also assume that $X$ is continuously embedded in $Y$ and $X$ is reflexive.
If $x_n$ is a sequence in $X$, weakly converging to $y \in Y$, then we must have: $$ y\in X$$ and $x_n$ converge weakly in $X$ to $y$.
This is false. Take, for example $L^3(0, 1)\subset L^2(0,1)$, $y\in L^2\setminus L^3$ and $x_n\in C^\infty(0,1)$ such that $x_n\to x$.
If, however, $x_n$ is bounded in $X$, then it is true. Indeed, since $X$ is reflexive, there is a convergent subsequence $x_{k(n)}$. Now, by definition, $x_{k(n)}\to x$ for some $x\in X$, but by assumption $x_{k(n)}\to y$, therefore $y=x$ and we can conclude that $y\in X$.
It remains to show that $\langle x_n-x , x^\star\rangle \to 0$ for all $x^\star \in X^\star$. Since $X\subset Y$, it holds that $Y^\star \subset X^\star$. Here I have to assume that $Y^\star $ is dense in $X^\star$, which in practice is always the case. By assumption, $$ \langle x_n-x, y^\star\rangle \to 0, \quad \forall y^\star\in Y^\star.$$ Now, $$ |\langle x_n-x, x^\star\rangle \le |\langle x_n-x, x^\star - y^\star\rangle | + |\langle x_n-x, y^\star\rangle|, $$ and the right-hand side can be made as small as wanted by using the boundedness of $x_n$ and the denseness of the embedding $Y^\star\subset X^\star$.