If $X\subset Y$ then $X^*\subset Y^*$ Is the following true,
If $X$ and $Y$ are Banach spaces and $X\subset Y$, then $X^*\subset Y^*$.
One argument for this is the following let $i:X\to Y$ be the identity map which implies its one to one and hence $i^* : Y^*\to X^*$ is onto which implies $X^* \subset Y^*$.
 Thank you 
 A: Actually, the reverse statement is (more or less) true.
I assume that by $X \subset Y$, you mean that the inclusion
$$
\iota : X \to Y, x \mapsto x
$$
is a well-defined, bounded linear map.
Then, for each $\varphi \in Y^\ast$, we get that $\varphi|_X = \varphi \circ \iota \in X^{\ast}$.
If also $X \subset Y$ is dense, then the "inclusion" map
$$
\Gamma : Y^\ast \to X^\ast , \varphi \mapsto \varphi|_X
$$
is also injective (why?), so that we can consider $Y^\ast$ as a subspace of $X^\ast$, using $\Gamma$ as the identification.
Your inclusion does not hold in general. To see this, consider the inclusion $X := C^1([a,b]) \subset C^0 ([a,b]) =: Y$, where we equip these spaces with the usual norms. Then
$$
\varphi : C^1([a,b]) \to \Bbb{C}, f \mapsto f'((a+b)/2)
$$
is a bounded linear functional on $C^1$, but it is easy to see that it has no continuous extension to $C^0$, because $\varphi$ is not bounded w.r.t. the $C^0$-norm (why?).
Hence, $\varphi \in X^\ast$, but $\varphi \notin Y^\ast$ (in any reasonable sense), so that $X^\ast \subset Y^\ast$ does not hold.
