$E$ be a $n+1$ dimensional vector space of functions on a domain $D$. Let $x_0,x_1,\ldots,x_n$ are distinct points in $D$. Show that the interpolation problem :
Find $f\in E$ such that $f(x_i)=y_i$ for $0\leq i\leq n$
has a unique solution for any choice of $y_0,y_1,\ldots,y_n$ if and only if no element of $E$ other that $0$ vanishes at all the points $x_0,x_1,\ldots,x_n$.
My Idea : I proved this problem for when $E$ is the space of polynomials of degree at most $n$. Now since any two vector spaces of dim- $n+1$ is isomorphic we get a isomorphism $\phi : E\to \mathbb{P_n}$ and the problem then transfered to the new one as
find $\phi(f)=p\in\mathbb{P_n}$ such that $p(x_i)=\phi(y_i)$ for $0\leq i\leq n$
has unique soln for any choice of $y_i$ if and only if no element of $\mathbb{P_n}$ other that zero poly vanishes at $x_1,...,x_n$. Since this new problem I know so taking $\phi^{-1}$ I got the result. Is there any flaw in this argument ? Is it rigorous?
Your proof is incorrect. Yes, there exists an isomorphism $\phi:E\to P_n$, but this doesn't tell you that the translated problem is to find $p$ such that $p(x_i)=y_i$. Instead, you want to find $p$ such that $f(x_i)=y_i$ where $f=\phi^{-1}(p)$. This isn't helpful, since you don't know anything about how the values of $f$ are related to the values of $p$.
What I would suggest instead is to imitate the method of proof you used to prove the result in the case $E=P_n$. In other words, don't use an isomorphism $\phi:E\to P_n$, but instead just try carrying out the same steps but with $E$ in place of $P_n$.