something to do with linear transformations $T$ is a linear map from $U$ to $V$, $S$ is a linear map from $V$ to $W$. $U$ and $V$ are finite dimensional Vector Spaces. We need to show that $\dim \ker ST\le \dim \ker S + 
\dim \ker T$.
Here is how I proceeded. Since $U,V $ are finite dimensional I can use the rank-nullity theorem on them which gives me $\dim V= \dim \ker S+ \dim {\cal R} S$, $\dim U=\dim \ker T + \dim {\cal R} T$and $\dim U=\dim \ker ST+\dim {\cal R}(ST)$. 
Now $\dim {\cal R} T\le \dim V$, so $\dim U\le \dim \ker T + \dim V$ which gives $\dim \ker (ST)+\dim {\cal R} (ST) \le \dim \ker T+\dim \ker S+ \dim {\cal R} S$. This is where I am stuck. I am unable to show any relationship between $\dim {\cal R} (ST)$ and  $\dim {\cal R} S$.
 A: Hint: For the relationship between $\dim\mathcal{R}(ST)$ and $\dim\mathcal{R}S$ note that $$\{S(T(x)) \mid x \in U\} = \{S(y) \mid y \in \mathcal{R}T\} \subseteq \{S(y) \mid y \in V\}.$$ I'm not sure this will help you with your current attempt at a proof though.
A: Hint: If we denote $X = TU\cap\operatorname{null}S\subset V$ (a subspace), it is fairly straightforward to show that
$$\dim\operatorname{null}ST = \dim\operatorname{null}T + \dim X.$$
A: I find working directly with dimension inequalities unsatisfying.
First note that the subspace of interest is $\ker (ST)$, and note that we must have $\ker T \subset \ker (ST)$. This suggests 'splitting' $\ker (ST)$ into two parts, $\ker T$ and the 'remainder'.
Let $u_1,...,u_k$ be a basis for $\ker T$. That is $\ker T = \operatorname{sp} \{ u_i \}$ (and $k = \dim \ker T$, of course). Now add vectors to turn this into a basis for $\ker (ST)$, let these vectors be $v_1,...,v_p$. Note that the vectors $u_1,...,u_k, v_1,...,v_p$ are linearly independent.
Then $\ker (ST) = \operatorname{sp} \{ u_1,...,u_k, v_1,...,v_p \}$, that is $\dim \ker (ST) = k+p = \dim \ker T + p$.
To finish, we need to show that $p \le \dim \ker S$.
The important step here is to note that $Tv_1,...,T v_p$ are linearly independent. (If not, then we would have some $\alpha \ne 0$ such that $\sum_i \alpha_i T v_i = T (\sum_i \alpha_i v_i) = 0$ which would mean that $\sum_i \alpha_i v_i \in \ker T$, and this in turn would mean that $\sum_i \alpha_i v_i$ could be written in terms of $u_1,...,u_k$ which would contradict our construction.)
Since the $v_i \in \ker (ST)$, we have $T v_1,...,T v_p \in \ker S$, and since they are linearly independent, we must have $p \le \dim \ker S$, which finishes our proof.
