Find a basis of $U=\{ p \in \mathcal{P}_4(\mathbb{F}) \ | \ \int_{-1}^{1} p = 0 \}$.
My method behind showing my list of vectors spans the space seemed a bit odd, so I was hoping if I could get a second pair of eyes to glance over things.
Attempt:
Let's let $p(x) \in U$ be arbitrary. Note then then that $p$ is of the form $$p(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0.$$ Further, we have \begin{align*} \int_{-1}^{1} \sum_{i=0}^4a_ix^i = 0 &\implies \sum_{i=0}^4 \frac{a_i}{i+1}x^{i+1} \Bigg|_{-1}^{1} = 0 \\ &\implies \frac{2a_4}{5}+\frac{2a_2}{3}+2a_0=0. & \qquad (*) \end{align*} Thus it's clear that $a_1,a_3$ can take on any value, hence $x,x^3 \in U$. If we solve $(*)$ for $a_0$, we see that $x^4+x^2-\frac{16}{30}\in U$. And if we set $a_0=0$, we also obtain that $-\frac{3}{5}x^4+x^2 \in U$.
We now show the obtained list of vectors is linearly independent. Take $a_1, \dots, a_4 \in \mathbb{F}$. Then from $$a_1x+a_2x^3+a_3 \left( x^4+x^2-\frac{16}{30}\right)+a_4 \left( -\frac{3}{5}x^4+x^2 \right)=0$$ we have $$a_1x+(a_3+a_4)x^2+a_2x^3+(a_3-\frac{3}{5}a_4)x^4=0$$ therefore $$\begin{cases} a_1=0 \\ a_3+a_4=0 \\ a_2=0 \\ a_3-\frac{3}{5}a_4=0 \end{cases}$$ which shows that $a_1=\cdots=a_4=0$, so the list is linearly independent.
Therefore we know that $\dim U \geq 4$. By a previous theorem, we must have $\dim U \leq \dim \mathcal{P}_4(\mathbb{F})=5$. But by one of the previous exercises, for a subspace $U$ of $V$, if $\dim U = \dim V$ then $U=V$, and here we have $\frac{1}{2} \in \mathcal{P}_4(\mathbb{F})$ but $\frac{1}{2} \not \in U$, so this cannot be the case. Thus $\dim U=4$ and the list of vectors must span the space.
An easier way to do this problem is to use the isomorphism between $\mathcal{P}_4(\mathbb{F})$ and $\mathbb{F}^5$, which you already constructed. Then the condition reads $2a_0+2a_2/3+2a_4/5=0$ in $\mathbb{F}^5$, as you computed. To construct a basis for the solution space to this homogeneous linear "system" (of one equation), rearrange it to have one of the three variables on one side and the others on the other side. The first variable will be your single pivot variable, and the other four variables are free.
The "standard recipe" for constructing a basis is then to set each free variable equal to $1$ and all others equal to $0$ and obtain a corresponding value of the pivot variable(s). This bypasses this awkward argument about dimensions, which would be much more annoying to write if you had, say, two independent linear equations. This choice of $1$s and $0$s is not essential, any sequence of vectors which can be assembled into an invertible matrix would suffice. The identity matrix is just a nice simple choice.
Then to get a basis back in the original space, just use the isomorphism to go back.