Why can an element in a vector space with an infinite basis always be in the span of a finite set? I'm working on a practice problem that says the following:

Let $V$ be an $F$-vector space and $X, Y$ infinite bases for $V$. Show that for any $x \in X$ you can find a finite $Y_0 \subset Y$ with $x$ in the subspace generated by $Y_0$.

Here's my confusion:
Let $V$ be the space of all polynomials in one variable $t$. I.e. $V = <1,t,t^2,t^3,...>$. Then 
\begin{align*}
 X &= \{1+t+t^2+... , t, t^2, t^3, t^4, ...\} \\
Y&= \{1, t, t^2, t^3, ...\}
\end{align*}
are both definitely bases for $V$. But for $x := 1+t+t^2 + ...$ there is definitely not a finite subset of $Y$ with $x$ in its span.
What's wrong with my thinking here?

When you have a look at the proof of the fact that every theorem has a basis you see that every element can be written as a finite sum of basis vectors. So let's have a look at the outline of the proof.
Let $P:=\{S\subset V | S \text{ is linearly independent}\}$ be the set of all linearly independent subsets with partial order "$\subset$". Using the Lemma of Zorn it is possible to prove that there exists a maximal element $M$ in $P$. Now $span_F M=V$, because otherwise there would exist some $x \in V$ which can NOT be written as the finite sum of elements of $M$. But then $M\cup \{x\}$ would be linearly independent because if
\begin{equation}\lambda_0 x+\lambda_1 v_1+...+\lambda_n v_n=0,
\end{equation} then either $\lambda\neq 0$, which by rearranging and the fact that $F$ is a field would imply that $x$ is in the span of $M$, but this is a contradiction. So $\lambda_0=0$, which would then imply $\lambda_i=0$ for all $i$ (because the $v_i$ are linearly independent). So $M\cup \{x\}$ is linearly independent. But this contradicts the maximality of $M$.
So you see that every element of $V$ can be written as a finite sum of elements of a certain set $M$ (because linearly independence is defined via finite sums).
