Is the following argument correct?
Suppose $U$ and $V$ are finite-dimensional vector spaces and $S\in\mathcal{L}(V,W)$ and $T\in\mathcal{L}(U,V)$. Prove that $$\dim \operatorname{null}ST\leq \dim\operatorname{null}S+\dim\operatorname{null}T$$
Proof. Prior to addressing the claim in question we first prove the ensuing proposition. $$\dim\operatorname{null}ST = \dim\operatorname{null}T+\dim\operatorname{range}T\cap\operatorname{null}S$$ Let $u_1,\dots,u_m$ and $Tx_1,\dots,Tx_n$ be bases for $\operatorname{null}T$ and $\operatorname{range}T\cap\operatorname{null}S$ respectively. We show that $u_1,\dots,u_m,x_1,\dots,x_n$ is a basis for $\dim\operatorname{null}ST$.
So assume that $\sum_{i=1}^{m}\lambda_iu_i+\sum_{i=1}^{n}\alpha_ix_i = 0$ where $\lambda_1,\dots,\lambda_m,\alpha_1,\dots,\alpha_n$ are arbitrary scalars in $\mathbf{F}$, applying $T$, yields $T\left(\sum_{i=1}^{m}\lambda_iu_i+\sum_{i=1}^{n}\alpha_ix_i\right) = \sum_{i=1}^{m}\lambda_iTu_i+\sum_{i=1}^{n}\alpha_iTx_i= \sum_{i=1}^{n}\alpha_iTx_i = 0$, then since $Tx_1,\dots,Tx_n$ is a basis for $\operatorname{range}T\cap\operatorname{null}S$, it follows that $\alpha_1 = \cdots = \alpha_n = 0$, consequently $\sum_{i=1}^{m}\lambda_iu_i= 0$ and since $u_1,\dots,u_m$ is a basis for $\operatorname{null}T$ it follows that $\lambda_1 =\cdots=\lambda_n = 0$, consequently the above list is linearily independent.
Now before showing that $\operatorname{null}ST = \operatorname{span}(u_1,\dots,u_m,x_1,\dots,x_n)$. We first demonstrate that $\operatorname{null}T\cap\operatorname{span}(x_1,x_2,\dots,x_n) = \{0\}$, Assume $v\in\operatorname{null}T\cap\operatorname{span}(x_1,x_2,\dots,x_n)$, then $v = \sum_{k=1}^{n}\rho_kx_k$ and $Tv = 0$, and by extension $T( \sum_{k=1}^{n}\rho_kx_k) = \sum_{k=1}^{n}\rho_kTx_k = 0$, a final appeal to $x_1,\dots,x_n$ being a basis of $\operatorname{range}T\cap\operatorname{null}S$, implies that $p_1 = \cdots = p_n = 0$ and so $v = 0$.
Now let $v\in\operatorname{null}ST$, either $v\in\operatorname{null}T$ or $v\in\operatorname{range}T\cap\operatorname{null}S$, in the event of the former we have $v = \sum_{i=1}^{m}\lambda_iu_i$ for some $\lambda_1,\dots,\lambda_n\in\mathbf{F}$ and in the case of the latter we have $v = \sum_{j=1}^{n}\alpha_jTx_j$ for some $\alpha_1,\dots,\alpha_n\in\mathbf{F}$, demonstrating that $\operatorname{null}ST = \operatorname{span}(u_1,\dots,u_m,x_1,\dots,x_n)$.
We may now affirm that $\dim\operatorname{null}ST = \dim\operatorname{null}T+\dim\operatorname{range}T\cap\operatorname{null}S$, observing that $\operatorname{range}T\cap\operatorname{null}S$ is a subspace of $ \operatorname{null}S$, theorem $\textbf{2.38}$, implies that $\dim\operatorname{range}T\cap\operatorname{null}S\leq\operatorname{null}S$, then $\dim\operatorname{null}ST = \dim\operatorname{null}T+\dim\operatorname{range}T\cap\operatorname{null}S\leq \dim\operatorname{null}T+\dim\operatorname{null}S$.
$\blacksquare$
NOTE - $2.38$ say that $\dim U\leq\dim V$ if $U$ is a subspace of $V$.