It could be a naive question. Probably, it is not true.
However, this question makes sense in the setting of function spaces.
For example, for $L\_\infty (0,1)$, we have $L\_p(0,1)\supset L\_\infty (0,1)$
and $L\_p(0,1)$ is reflexive when $p>1$.
On the other hand, $L\_1(0,1)$ itself is weakly sequentially complete and any rearrangement-invariant function spaces on $(0,1)$ is a subset of $L\_1(0,1)$.


I am wondering whether we have such a result in the setting of general Banach spaces.


BTW: Davis, Figiel, Johnson and Pełczyński's construction gives us a smaller reflexive space.



For a Banach space $X$, can we find a reflexive (or weakly sequentially complete) space $Y$ such that $X\subset Y$?
Let $K$ be a compact scattered space and $X=C(K)$ the space of continuous functions on $K$. We want to show that there is no injective bounded linear $T:X\to Y$ into a weakly sequentially complete (w.s.c.) Banach space $Y$ if $K$ has *large*(see below) cardinality.


Let $T:X\to Y$ be as above. $C(K)$ has Pelczynski property (V) and $Y$ is w.s.c., so $T$ is weakly compact. Since $C(K)$ has Dunford-Pettis property, then $T$ is completely continuous. Lastly $C(K)$ contains no copy of $\ell^1$ since $K$ is scattered. Thus, $T$ is compact.


Since $T$ is compact, its range $T(X)$ is separable, so $card(T(X))\leq\mathfrak{c}=card(\mathbb{R})$. $T$ is injective, so $card(X)=card(T(X))$. Together, we have
$$ card(K)\leq card(X)=card(T(X))\leq\mathfrak{c}.$$
However, this leads to a contradiction when $card(K)>\mathfrak{c}$. For example, take $K=[0,\alpha]$ for some ordinal $\alpha$ with the order topology, where the cardinality of $\alpha$ is strictly greater than $\mathfrak{c}$.




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Note (2023-02-18): Following the footsteps in this post, we can actually make a slightly more general statement. If $X$ is a Banach space with property (V), $X^\*$ has Schur property, and $card(X)>\mathfrak{c}$, then there exists no injective bounded $T:X\to Y$ into a w.s.c. $Y$.


