Proof check - infinite-dimensional $\mathfrak{sl}(2, \mathbb{F})$-module Let $L=\mathfrak{sl}(2,\mathbb{F})$ with the usual basis $(x, \ y, \ h)$ and $\text{char}\,\mathbb{F}=0$.
Let $Z(\lambda)$, $\lambda\in\mathbb{F}$ the infinite-dimensional $L$-module spanned by $\left(v_{0}, \ v_{1},\dots\right)$, $v_{i}=\frac{1}{i!}y^{i}\cdot v_{0}$ for which:
a) $h\cdot v_{i}=(\lambda-2i)v_{i}$,
b) $y\cdot v_{i}=(i+1)v_{i+1}$, and
c) $x\cdot v_{i}=(\lambda-i+1)v_{i-1}$ (where $v_{-1}=0$).
I was trying to proof the two statements:
I) Every submodule has a maximal vector.
II) If $\lambda+1$ is not a positive integer, then $Z(\lambda)$ is irreducible.
I tried the following:
I) My problem lies with the infinite-dimensional submodules:
First, let $U\subseteq Z(\lambda)$ be an infinite-dimensional submodule and $0\neq u\in U$ arbitrary. Write $u=\sum\nolimits_{i=0}^{t}\lambda_{k_{i}}v_{k_{i}}$ so that all $\lambda_{k_{i}}\neq 0$. It is $h\cdot u=\sum\nolimits_{i=0}^{t}\lambda_{k_{i}}(\lambda-2k_{i})v_{k_{i}}$ so $\color{red}{h\text{ acts diagonally on  }U\text{ and  therefore }v_{k_{i}}\in U}$. So $U$ is the direct sum of $\mathbb{F}v_{j}$, $j\in J\subseteq\mathbb{N}$. Let $r$ be minimal for $v_{r}\in U$, so $v_{r-1}\notin U$. But since $U$ is a submodule $x\cdot v_{r}$ has to be. This means $x\cdot v_{r}=0$ and $v_{r}$ is a maximal vector.
II) Suppose $Z(\lambda)$ is not irreducible, than a submodule $U\neq0$ exists. By the preceding paragraph, it has a maximal vector. $\color{red}{\text{It  has  the form }v_{k}}$, where $k>0$, since otherwise $U$ would be all of $Z(\lambda)$. That means it is: $0=x\cdot v_{k}=(\lambda-k+1)v_{k-1}$ since $v_{k-1}\neq 0$ it is $\lambda-k+1=0\iff \lambda+1=k$. This contradicts the choice of $\lambda$.
I have highlighted the parts that I think that could be problematic. For the first one I am not sure if it is that easy and for the second if the maximal vector really has to have that form.  
Thank you for helping me.
 A: I.  I think you are trying to show that $U$ is also a weight $L$-module.  However, I'm not sure if you understand how "$h$ acts diagonally on $U$" would imply $v_{k_i}\in U$.  Do you know how to prove this?  (I can give you hints, but I'm not sure you need them.)  You are definitely right that $U$ is the direct sum of $\mathbb{F}v_j$ whenever $v_j$ occurs as a summand in an element of $U$.
In addition, I'm not sure what you meant by "But since $U$ is a submodule, $x\cdot v_r$ has to be."  Can you clarify on that?  Nonetheless, the conclusion is correct.
II.  You might want to say "a proper submodule $U\neq 0$ exists."  The red part is better written as "This maximal vector may be taken to be $v_k$ for some $k>0$."  Since all of your weight spaces are either $0$- or $1$-dimensional, you are safe.  The same cannot be said for, for example, representations of $\mathfrak{sl}_3$.

Non-Inductive Proof of the Red Text in I.
Let $\mu_i$ be the weight of $v_{k_i}$ (i.e., $\mu_i=\lambda-2k_i$).  Then, consider the element 
$$h_i:=\left(h-\mu_1\right)\left(h-\mu_2\right)\cdots\left(h-\mu_{i-1}\right)\left(h-\mu_{i+1}\right)\cdots \left(h-\mu_t\right)$$
in the universal enveloping algebra $\mathfrak{U}(L)$ of $L$.  Clearly, $h_i\cdot u \in U$.  However,
$$h_i\cdot u=\Bigg(\lambda_{k_i}\,\prod_{j\neq i}\,\left(\mu_i-\mu_j\right)\Bigg)\,v_{k_i}\,.$$
Therefore, $\Bigg(\lambda_{k_i}\,\prod_{j\neq i}\,\left(\mu_i-\mu_j\right)\Bigg)\,v_{k_i}\in U$.  Since $\lambda_{k_i}\,\prod_{j\neq i}\,\left(\mu_i-\mu_j\right)\neq 0$, $v_{k_i}\in U$.
