If $B\subset A$ and the exist a injection $A\rightarrow B$, then $A$ and $B$ has same cardinality? By using the definition, the cardinality is the same iff there is a bijection. So if $B\subset A$ and the exist a injection $f\colon A\rightarrow B$, then $A$ and $B$ has same cardinality.
 A: Any attempt to prove this result without explicitly using the Cantor–Bernstein–Schroeder Theorem will, in effect, end up re-proving the aforementioned theorem.  
Let us call your statement "mini-CBS": Whenever $B \subseteq A$ are sets such that there is an injection $f : A \to B$, then $\left| A \right| = \left| B \right|$.
We now prove the CBS Theorem from mini-CBS: Suppose that $X$ and $Y$ are sets, and we have injections $f: X \to Y$ and $g : Y \to X$.  Consider the set $X^\prime = \{ g(y) : y \in Y \}$.  Clearly, $X^\prime \subseteq X$.  Note that $g$ is in fact a bijection between $Y$ and $X^\prime$, and so $\left| X^\prime \right| = \left| Y \right|$.  Consider now the composition $g \circ f : X \to X^\prime$.  Since $f$ and $g$ are both one-to-one, it follows that $g \circ f$ is also one-to-one.  By mini-CBS it follows that $\left| X^\prime \right| = \left| X \right|$, and therefore $\left| Y \right| = \left| X \right|$.
A: Yes that's right. $B\subset A$ implies that $|B|\le |A|$. The injection forces $|A|\le|B|$. Therefore they have the same cardinality.
A: If $B \subset A$ then there exists an injection $g : B \to A$ defined in the natural way (identity). By existence of $g$ and the injection $f : A \to B,$ invoke Cantor–Bernstein–Schroeder theorem then there exist a bijection $h : A \to B.$ Hence $A$ and $B$ has the same cardinality.
