Norm and Weak Topologies agree?

We can identify our space $X$ with $\mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $\mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_\alpha\to x$ weakly, then $f_i(x_\alpha)\to f_i(x)$ where $f_i:\mathbb{R}^n\to\mathbb{R}$ is the $i$th coordinate functional. But that just means that $x_\alpha$ converges to $x$ on each coordinate, which implies $x_\alpha$ converges to $x$ in the Euclidean topology. Thus $x_\alpha\to x$ in the norm topology.

More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance

Another way to do this is to just go back to the definitions. Let $\epsilon>0,$ and $x_0\in X$. It suffices to show that any open ball $B(x_0,\epsilon)$ is weakly open. We may use $\|\cdot\|_{\infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=\sum^n_{i=1} x_ie_i$ for each $x\in X$ and for each $1\le i\le n,$ we have the functionals $e_i^*:X\to \mathbb C:x\mapsto x_i.$

Then, $B_{\epsilon }=\left \{ x:\left \| x\right \|<\epsilon \right \}=\left \{ x:\forall\ 1\le i\le n,\ |x_i|<\epsilon \right \},\ $ which is nothing more than the weakly open set $\left \{ x:\forall\ 1\le i\le n,\ |e_i^*(x)|<\epsilon \right \}.$