A weak to norm continuous operator between Banach spaces is a finite rank operator Let $X$,$Y$ be Banach spaces. And $T:X\rightarrow Y$ a continuous operator when $X$ is endowed with the weak topology and $Y$ with the norm topology. I am trying to show that $T$ is finite rank i.e. that its range is a finite dimensional vector space.
The first thing I tried was taking a basis in $X$ and looking at the image but I do not think the assumptions say anything useful about this. 
I also saw on some other post the same question for Hilbert spaces in which they looked at the inverse image of the (closed) unit ball, but I am not sure how they arrived at that idea and how I can use this. 
A subset $U$ of $X$ is open if there exist a $\rho>0$ and $\omega_1,\ldots,\omega_n\in X^*$ the dual space of $X$ such that $\{y\in X: \lvert\omega_i(x-y)\rvert<\rho\}\subset U $ for all $\omega_i$.
 A: By continuity at $0$ there exist $n,w_1,w_2,...,w_n$ and $r_1,r_2,...,r_n$ such that $|w_i(x)| <r_i, 1\leq i \leq n$ implies $\|Tx\|<1$. From this it is easy to see that $w_i(x)=0, 1\leq i \leq n$ implies $Tx=0$ (Just consider integer multiples of $x$).
Now consider  the quotient space $X/ker(T)$. The map $x+\ker(T) \to (w_1(x),w_2(x),..., w_n(x))$ is a one-to-one linear map from $X/ker(T)$ to $\mathbb R^{n}$ so the dimension of $X/ker(T)$ is at most $n$. 
If $x_j+\ker (T), 1 \leq j  \leq N$ is  basis for this space then $x \in X$ implies $x+\ker (T) =\sum a_i(x_i+\ker (T))$ for some scalars $a_i$ which shows that $Tx$ belongs to the space spanned by $Tx_1,Tx_2,...,Tx_n$. 
A: We know that the inverse image of $B_Y(0,1)$ contains an open subset, so there are linear forms $u_1,\ldots,u_n$ on $X$ and some $\epsilon > 0$ such that for any $x \in X$, if $|u_i(x)| < \epsilon$, then $\|Tx\| < 1$. 
In particular, let $K$ be the intersection of the kernels of the $u_i$. Then $T_{|K}$ is a linear map that is bounded (as in, a bounded function), so is zero. 
So, it remains to show that if $T:X \rightarrow Y$ is linear and vanishes on an intersection of finitely many hyperplanes, then $T$ has finite rank. 
We prove it by assuming that $T$ vanishes on the intersections of the kernels of the $v_i$ ($1 \leq i \leq m$), $v_i \in X^*$, the algebraic dual. 
We can suppose that $m$ is minimal with this propriety, ie that for any $i$, $T$ doesn’t vanish on the intersections of the kernels of the $v_j$, $j \neq i$. 
Then let, for any $1 \leq i \leq m$ be $y_i$ such that $v_j(y_i)=0$ ($j \neq i$) and $T(y_i) \neq 0$, thus $v_i(y_i)\neq 0$ (up to scaling by a constant factor we may assume $v_i(y_i)=1$). 
Let now $x \in X$. Then it is easy to check that $x-\sum_{i=1}^m{v_i(x)y_i}$ is in the kernel of every $v_i$, so in the kernel of $T$, thus $T(x)$ is a linear combination of $T(y_1),\ldots,T(y_n)$ and we are done. 
