Let A is a C*-algebra, A$^+$ means A$\times$$\mathbb{C}$ equipped with pointwise sum and with a multiplication defined by:
(a, $\lambda$)(b, $\mu$) = (ab + $\lambda$b + $\mu$a, $\lambda$$\mu$)
If x, y $\in$ A are homotopy in some subset of A$^+$, for example the set of unitary elements, do they must be homotopy in A?
In general, if A is a sub-C*-algebra of C*-algebra B, if x, y $\in$ A are homotopy in some subset of B, do they must be homotopy in A?
If that's your definition, then every $x,y\in A$ are in homotopy, in any C$^*$-algebra: you just take $$ \gamma(t)=(1-t)x+ty,\ \ \ t\in[0,1]. $$
The answer to your last question is no: let $A=\mathbb C\oplus\mathbb C\subset M_2(\mathbb C)$, represented as $$ A=\left\{\begin{bmatrix}a&0\\0&b\end{bmatrix}:\ a,b\in\mathbb C\right\}\subset M_2(\mathbb C), $$ and consider homotopy in the corresponding sets of projections. Let $$ p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ q=\begin{bmatrix}0&0\\0&1\end{bmatrix}. $$ Then $p$ and $q$ are homotopic via projections in $M_2(\mathbb C)$, by $$ \gamma(t)=\begin{bmatrix}t&\sqrt{t-t^2}\\ \sqrt{t-t^2}&1-t\end{bmatrix},\ \ t\in[0,1], $$ but they are not homotopic by projections in $A$.
With the same algebras, and unitaries: $$ u=\begin{bmatrix}1&0\\0&-1\end{bmatrix},\ \ v=\begin{bmatrix}-1&0\\0&1\end{bmatrix}. $$ are homotopic via unitaries in $M_2(\mathbb C)$, by $$ \gamma(t)=\begin{bmatrix}\cos t\pi&\sin t\pi \\ \sin t\pi&-\cos t\pi\end{bmatrix},\ \ t\in[0,1], $$ but they are not homotopic by unitaries in $A$.