Of course the dimension is an obvious obstacle, but even if the space have the same cardinality of Hamel bases the answer is no. For example in the paper

A. Avilés, P. Koszmider, A Banach space in which every injective operator is surjective. Bull. Lond. Math. Soc. 45 (2013), no. 5, 1065–1074

the authors constructed an infinitely dimensional Banach space $X$ such that if $T:X\to X$ is bounded and injective, then $T(X)=X$. Therefore is $Y$ is a subspace of $X$, then one cannot find an injective operator $T:X\to Y$.

Of course the dimension is an obvious obstacle, but even if the space have the same cardinality of Hamel bases the answer is no. For example in the paper

A. Avilés, P. Koszmider, A Banach space in which every injective operator is surjective. Bull. Lond. Math. Soc. 45 (2013), no. 5, 1065–1074

the authors constructed an infinitely dimensional Banach space $X$ such that if $T:X\to X$ is bounded and injective, then $T(X)=X$. Therefore if $Y$ is a subspace of $X$, then one cannot find an injective operator $T:X\to Y$.