I got it! Well, with help. I found the part of the book that gives steps to work through when proving the $Var \left( \hat{\beta}_0 \right)$ formula (thankfully it doesn't actually work them out, otherwise I'd be tempted to not actually do the proof). I proved each separate step, and I think it worked.
I'm using the book's notation, which is:
$$
SST_x = \displaystyle\sum\limits_{i=1}^n (x_i - \bar{x})^2,
$$
and $u_i$ is the error term.
1) Show that $\hat{\beta}_1$ can be written as $\hat{\beta}_1 = \beta_1 + \displaystyle\sum\limits_{i=1}^n w_i u_i$ where $w_i = \frac{d_i}{SST_x}$ and $d_i = x_i - \bar{x}$.
This was easy because we know that
\begin{align}
\hat{\beta}_1 &= \beta_1 + \frac{\displaystyle\sum\limits_{i=1}^n (x_i - \bar{x}) u_i}{SST_x} \\
&= \beta_1 + \displaystyle\sum\limits_{i=1}^n \frac{d_i}{SST_x} u_i \\
&= \beta_1 + \displaystyle\sum\limits_{i=1}^n w_i u_i
\end{align}
2) Use part 1, along with $\displaystyle\sum\limits_{i=1}^n w_i = 0$ to show that $\hat{\beta_1}$ and $\bar{u}$ are uncorrelated, i.e. show that $E[(\hat{\beta_1}-\beta_1) \bar{u}] = 0$.
\begin{align}
E[(\hat{\beta_1}-\beta_1) \bar{u}] &= E[\bar{u}\displaystyle\sum\limits_{i=1}^n w_i u_i] \\
&=\displaystyle\sum\limits_{i=1}^n E[w_i \bar{u} u_i] \\
&=\displaystyle\sum\limits_{i=1}^n w_i E[\bar{u} u_i] \\
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i E\left(u_i\displaystyle\sum\limits_{j=1}^n u_j\right) \\
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \left[E\left(u_i u_1\right) +\cdots + E(u_i u_j) + \cdots+ E\left(u_i u_n \right)\right] \\
\end{align}
and because the $u$ are i.i.d., $E(u_i u_j) = E(u_i) E(u_j)$ when $ j \neq i$.
When $j = i$, $E(u_i u_j) = E(u_i^2)$, so we have:
\begin{align}
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \left[E(u_i) E(u_1) +\cdots + E(u_i^2) + \cdots + E(u_i) E(u_n)\right] \\
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i E(u_i^2) \\
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \left[Var(u_i) + E(u_i) E(u_i)\right] \\
&= \frac{1}{n}\displaystyle\sum\limits_{i=1}^n w_i \sigma^2 \\
&= \frac{\sigma^2}{n}\displaystyle\sum\limits_{i=1}^n w_i \\
&= \frac{\sigma^2}{n \cdot SST_x}\displaystyle\sum\limits_{i=1}^n (x_i - \bar{x}) \\
&= \frac{\sigma^2}{n \cdot SST_x} \left(0\right)
&= 0
\end{align}
3) Show that $\hat{\beta_0}$ can be written as $\hat{\beta_0} = \beta_0 + \bar{u} - \bar{x}(\hat{\beta_1} - \beta_1)$. This seemed pretty easy too:
\begin{align}
\hat{\beta_0} &= \bar{y} - \hat{\beta_1} \bar{x} \\
&= (\beta_0 + \beta_1 \bar{x} + \bar{u}) - \hat{\beta_1} \bar{x} \\
&= \beta_0 + \bar{u} - \bar{x}(\hat{\beta_1} - \beta_1).
\end{align}
4) Use parts 2 and 3 to show that $Var(\hat{\beta_0}) = \frac{\sigma^2}{n} + \frac{\sigma^2 (\bar{x}) ^2} {SST_x}$:
\begin{align}
Var(\hat{\beta_0}) &= Var(\beta_0 + \bar{u} - \bar{x}(\hat{\beta_1} - \beta_1)) \\
&= Var(\bar{u}) + (-\bar{x})^2 Var(\hat{\beta_1} - \beta_1) \\
&= \frac{\sigma^2}{n} + (\bar{x})^2 Var(\hat{\beta_1}) \\
&= \frac{\sigma^2}{n} + \frac{\sigma^2 (\bar{x}) ^2} {SST_x}.
\end{align}
I believe this all works because since we provided that $\bar{u}$ and $\hat{\beta_1} - \beta_1$ are uncorrelated, the covariance between them is zero, so the variance of the sum is the sum of the variance. $\beta_0$ is just a constant, so it drops out, as does $\beta_1$ later in the calculations.
5) Use algebra and the fact that $\frac{SST_x}{n} = \frac{1}{n} \displaystyle\sum\limits_{i=1}^n x_i^2 - (\bar{x})^2$:
\begin{align}
Var(\hat{\beta_0}) &= \frac{\sigma^2}{n} + \frac{\sigma^2 (\bar{x}) ^2} {SST_x} \\
&= \frac{\sigma^2 SST_x}{SST_x n} + \frac{\sigma^2 (\bar{x})^2}{SST_x} \\
&= \frac{\sigma^2}{SST_x} \left( \frac{1}{n} \displaystyle\sum\limits_{i=1}^n x_i^2 - (\bar{x})^2 \right) + \frac{\sigma^2 (\bar{x})^2}{SST_x} \\
&= \frac{\sigma^2 n^{-1} \displaystyle\sum\limits_{i=1}^n x_i^2}{SST_x}
\end{align}
