If $V/F$ has basis $\{x_1,x_2,x_3,\dotsc\}$, $T'\colon V \to V$ is the right-shift operator, $W\subset V$ is $T'$-invariant, then $\dim V/W < \infty$? I am studying from Jacobson's Basic Algebra II (2nd edition), and I am stuck on Exercise 3 of $\S$3.2 (on page 103).

Exercise 3.2.3. Let $V$ be a vector space over a field $F$, $T$ a linear transformation in $V$ over $F$, and let $F[\lambda]$ be the polynomial ring over $F$ in an indeterminate $\lambda$. Then $V$ becomes an $F[\lambda]$-module if we define
  $$
(a_0 + a_1 \lambda + a_2 \lambda^2 + \dotsb + a_m \lambda^m)x = a_0 x + a_1 (Tx) + a_2 (T^2 x) + \dotsb + a_m(T^m x).
$$
  Let $V$ have the countable base $(x_1,x_2,\dotsc)$ and let $T$ be the linear transformation such that $Tx_1 = 0$, $Tx_{i+1} = x_i$, $i = 1,2,3,\dotsc$. Show that $V$ as an $F[\lambda]$-module defined by $T$ is artinian. Let $T'$ be the linear transformation in $V$ such that $Tx_i = x_{i+1}$, $i = 1,2,\dotsc$. Show that $V$ as an $F[\lambda]$-module defined by $T'$ is noetherian.

My solution.
I am able to show that $V$ as an $F[\lambda]$-module defined by the left-shift operator $T$ is artinian as follows. Firstly, the $F[\lambda]$-submodules are precisely the $T$-invariant subspaces of $V$. I showed that any $T$-invariant subspace of $V$ is either finite-dimensional or equal to the full space $V$. So, any proper descending chain of $T$-invariant subspaces of $V$ must contain a finite-dimensional subspace $W$, which forces the chain to terminate from that point onwards in no more than $\dim W$ steps. Thus $V$ is artinian as an $F[\lambda]$-module defined by $T$.
Now, to show that $V$ is noetherian as an $F[\lambda]$-module defined by the right-shift operator $T'$, as before the $F[\lambda]$-submodules of $V$ are precisely the $T'$-invariant subspaces of $V$. Any nonzero $T'$-invariant subspace is necessarily infinite-dimensional, but I believe even more is true:

Claim: If $W$ is a nonzero $T'$-invariant subspace of $V$, then $W$ is of finite codimension, that is, $\dim V/W < \infty$.

If the claim is true, then I can complete the proof as follows: if $W_1 \subsetneq W_2 \subsetneq W_3 \subsetneq \dotsb$ is a proper increasing chain of $T'$-invariant subspaces of $V$, then we get a sequence of injective maps $V/W_1 \hookleftarrow V/W_2 \hookleftarrow V/W_3 \hookleftarrow \dotsb$. This sequence must terminate at $0$ in finitely many steps because $\dim V/W_1 < \infty$ and the map $V/W_{i+1} \hookrightarrow V/W_i$ is not surjective for any $i$. Hence, $W_i = V$ for some $i$, so $V$ is noetherian as an $F[\lambda]$-module defined by $T'$.

I am stuck in proving the above claim.
Here is how I proved that any $T$-invariant subspace of $V$ must be either finite-dimensional or equal to the full space $V$. Suppose that $W$ is an infinite-dimensional $T$-invariant subsapce of $V$. Let $w \in W$ be a vector whose projection onto the subspace spanned by $x_i$ is nonzero, for some $i \in \mathbb{N}$. Then, $T^{i-1}w$ is a scalar multiple of $x_i$, so $x_i \in W$. Then, $T^{i-2}w$ is a linear combination of $x_i$ and $x_{i-1}$, which combined with the previous fact shows that $x_{i-2}$ lies in $W$. Proceeding inductively, we get that $x_j \in W$ for all $j = 1,2,\dotsc,i$. Now, since $W$ is infinite-dimensional, there are infinitely many $i \in \mathbb{N}$ such that the projection of $W$ onto $Fx_i$ is nonzero. Hence, $x_j \in W$ for all $j = 1,2,\dotsc,i$ for infinitely many $i$, which implies that $x_j \in W$ for all $j \in \mathbb{N}$. Thus, $W = V$.
I am not able to "dualise" this argument for the case of the right-shift operator $T'$, but I strongly believe something along those lines should work to prove the claim. Perhaps I am missing only some small idea; I would appreciate any help on this.
 A: Let $W\subseteq V$ be a nonzero $T'$-invariant subspace and let $w=\sum a_i x_i$ be some nonzero element of $W$.  All but finitely many of the coefficients $a_i$ are $0$, so let $n$ be maximal such that $a_n\neq 0$.  Let $V_m$ be the subspace of $V$ spanned by $x_1,\dots,x_m$.  Observe that $V_n$ is in fact spanned by $x_1,\dots,x_{n-1},w$, since you can use a multiple of $w$ to get the desired coefficient of $x_n$ and then fix the coefficients of $x_i$ for each $i<n$.  Similarly, $V_{n+1}$ is spanned by $x_1,\dots,x_{n-1},w,T'w$: first use $T'w$ to get the correct coefficient of $x_{n+1}$, and then you just need to correct by an element of $V_n$ which we know we can do using $x_1,\dots,x_{n-1},w$.  In the same way, $V_{n+k}$ is spanned by $x_1,\dots,x_{n-1},w,T'w,\dots,{T'}^kw$ for any $k$, and so all of $V$ is spanned by $x_1,\dots,x_{n-1},w,T'w,\dots$.  But $w,T'w,\dots$ are all in $W$, so this means $x_1,\dots,x_{n-1}$ span $V$ modulo $W$, and so $V/W$ is finite-dimensional.
Alternatively, note that $V\cong F[\lambda]$ as an $F[\lambda]$-module, by mapping $x_i$ to $\lambda^{i-1}$.  Every ideal in $F[\lambda]$ is principal and in particular finitely generated, so $F[\lambda]$ is Noetherian.
(To connect the two approaches, the first argument is actually essentially the same as long division in $F[\lambda]$, showing that any element of $V$ can be "divided" by $w$ to get a remainder spanned by $x_1,\dots,x_{n-1}$.)
