Let $A$ be a C*-algebra. $\tilde{A}$ be the unitization of $A$. I checked the following lemma:
If $x$ and $y$ are elements of $M_n(\tilde{A})$ such that $x$ is invertible and $\|x-y\| \leq \frac{1}{\|x^{-1}\|}$, then the elements $x+t(y-x)$, $t \in [0,1]$ are all invertible in $M_n(\tilde{A})$.
Let $\pi:\tilde{A} \to \mathbb{C}$ be the map that takes $a+\lambda$ to $\lambda$ and it extends to $M_n(\tilde{A}) \to M_n({\mathbb{C}})$ by $(a_{ij}) \mapsto (\pi(a_{ij}))$ and $$GL_n^+(A)=\{a \in GL_n(\tilde{A})| \pi(a)=1\}$$ $$U_n^+(A)=\{a \in U_n(\tilde{A})| \pi(a)=1\}$$
Does it follow from this lemma that $GL_n^+(A)$ is open? And I know the information that $U_n(\tilde{A})$ is not open but I could not find a proper reason. Any help appreciated.
No, it doesn't. Even without a clear definition for $\pi$ (since it should act on $M_n(\tilde A)$), any reasonable definition of $\pi$ will be continuous, and so you can perturb an $a$ with $\pi(a)=1$ very slightly to get $\pi(a+\varepsilon)\ne1$. So $GL_n^+(A)$ is not open.
The unitary group, on the other hand, is closed (because a norm-limit of unitaries is a unitary), and now $$ U_n^+(A)=U_n(\tilde A)\cap \pi^{-1}(\{1\}). $$ is closed if $\pi$ is continuous.
You should consider improving your question a bit, in particular by writing a proper definition of $\pi$, and/or how it extends to $M_n(\tilde A)$.