The strong, weak and weak-star topologies coincide on finite dimensional spaces. Let $E$ be a finite dimensional normed linear space. I have been able to show using set inclusion that $s=w=w^*$, where $s,\,w$ and $w^*$ represent strong, weak and weak-star topologies, repsectively. 
However, the method I want to use in this case, is that the canonical map $J$ is surjective. So, how do I show it (I guess we would need to show, in the process, that $\dim E=\dim E^*=\dim E^{**}$).
The canonical map is defined as
\begin{align}J:&E\to E^{**}\\&x\mapsto J(x)=\phi_x \end{align}
where 
\begin{align}\phi_x:& E^{*}\to \mathbb{R}\\&f\mapsto \phi_x(f)=f(x) \end{align}
 A: There are two questions here: showing the weak topology coincides with the norm topology on a finite-dimensional space, and showing the weak$^*$ topology coincides with the weak topology on a finite-dimensional space.
The latter question essentially comes down to showing that finite-dimensional spaces are reflexive, meaning that $J$ is a surjective map. If you can show that $\dim E^* = \dim E$, then this pretty much does everything for you; you know that $\dim E^{**} = \dim E^* = \dim E$, and that $J$ is injective (as isometries always are), hence $J$ is surjective also.
You can also argue that the unit ball of $E$ is norm-compact by Heine-Borel, and hence is weakly-compact; a condition equivalent to reflexivity.
To show that $\dim E^* = \dim E$, you can fix a basis $x_1, \ldots, x_n \in E$, and for each $i$, define a linear map $f_i \in E^*$ by its action on the basis, particularly,
$$f_i(x_j) = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j.\end{cases}$$
Then, show that $f_i$s form a basis (in fact, showing they span $E^*$ suffices).
For the former question, I suggest constructing a bounded weakly open set. This cannot be done in an infinite-dimensional space. The advantage of a bounded weakly open set is that we can scale and translate it so that it contains $0$, and is contained in the open unit ball. Thus, every norm-open neighbourhood would also be a weakly-open neighbourhood, and the topologies would coincide.
Without loss of generality, take the norm to be the $\| \cdot \|_\infty$ norm on $\Bbb{R}^n$ (all norms are equivalent, so choosing this norm will produce the same norm and weak topology as any other norm). Then the unit ball of $\| \cdot \|_\infty$ is itself a basic weakly open set! If we define the $f_i$s, where $x_i$ is the $i$th standard basis vector, then we can express
$$B_{(\Bbb{R}^n, \| \cdot \|_\infty)}(0; 1) = \bigcap_{i=1}^n f_i^{-1}(-1, 1).$$
This should geometrically make sense; the $\infty$-ball is a big hypercube, which can be defined by the finitely many ($2n$) hyperplanes that support its facets.
Hence, the norm, weak, and weak$^*$ topologies all coincide.
