How can the little l2 topology be finer than the uniform topology? How can the $\ell_2$ topology be finer than the uniform topology on the set $X$ of square-summable sequences?
If the  $\ell_2$ metric is always greater than or equal to the uniform metric. Wouldn't this mean that the epsilon balls in  $\ell_2$ are always bigger than the epsilon balls under the uniform metric? I'm trying to show that the  $\ell_2$ balls are contained inside the uniform balls, right?

This is a misleading concept. Consider, for example, the real line. If you define the $\epsilon$-ball to be $B(x,\epsilon)=\{(y\space s.t.\space 2|y-x|<\epsilon\}$ this would make balls "2 times smaller", yet the topology is the same.
Remember, that even if for a fixed $\epsilon$ $d_1$-$\epsilon$-ball is smaller than the $d_2$-$\epsilon$-ball, there may still be a smaller $\delta$ such that $d_2$-$\delta$-ball is smaller than $d_1$-$\epsilon$-ball.
Now hint: to show that $\ell_2$ is finer than the uniform but coarser than the box topology consider $M(x)=\{y:|x_i-y_i|^2<\epsilon^2/2^i\}$ and show that $M(x)\subset B_{\ell_2}(x,\epsilon)\subset B_{\rho}(x,\epsilon)$ where $\rho$ is uniform metric.
