Suppose we have a transformation: $T(f(t))= (f(0),f(1),...) $ where $T:\mathbb{R}[t]\rightarrow V$ where $\mathbb{R}[t]$ is the set of real polynomials and $V$ is a vector space.
I think this is clearly an isomorphism. It's linear and we can find and inverse: $T^{-1}(f(0),f(1),...)= f(t)$. Clearly injective and surjective. Am I somehow wrong?
I assume that you mean $V$ is not just any vector space, but the vector space $\mathbb{R}^\mathbb{N}$ of infinite sequences of real numbers.
It's not an isomorphism because it isn't surjective. For example, there is no polynomial $f(t)$ such that $f(n) = 2^n$ for all $n$, which means that $$ (1, 2, 4, 8, 16, \ldots) \in V $$ is not in the image of $f$. In general, a polynomial of degree $n$ is determined by just the first $n+1$ coordinates, $f(0), f(1), f(2), \ldots, f(n)$ -- so by additionally providing $f(n+1), f(n+2),$ and so on you are over-specifying $f$, which explains intuitively why your map is not surjective.
It's linear and we can find and inverse: $T^{-1}(f(0),f(1),...)= f(t)$. Clearly injective and surjective. Am I somehow wrong?
By writing the argument to $T^{-1}$ as $(f(0), f(1), \ldots)$ you have just assumed that elements of $V$ have this form, which is only true if $T$ is surjective, so you are already assuming part of what you wanted to prove (the part that is actually false). Regardless, I don't exactly understand your description of $T^{-1}$.
In fact, $\mathbb{R}[t]$ is isomorphic to the vector space of infinite sequences of real numbers which are identically $0$ after a certain point. Maybe you can see why.