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This was one of the exercises in my lecture notes and I've absolutely no idea how to go about this.

Here both $S, T:V \rightarrow V$ and $\text{null }T := \dim(\ker T)$ where $\ker T$ is the set of vectors in $V$ that $T$ maps to $0$. (Just to clear up any possible confusion since I've seen different notation on here being used).

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Let us denote with $T'$ restriction of $T$ on $\ker ST$, i.e. $T'\colon \ker ST\to V$, $T'x = Tx$.

First, $\ker T' = \ker ST\cap \ker T$, but $\ker T\subseteq \ker ST$, so $\ker T' = \ker T$, or $\operatorname{null}T' = \operatorname{null}T$.

Secondly, $\operatorname{im} T'\subseteq \ker S$ because $ST'x = STx = 0$ for $x\in \ker ST$. That is, $\operatorname{rank} T'\leq \operatorname{null}S$.

Finally, by rank-nullity theorem we have $\operatorname{null}ST = \operatorname{rank}T'+\operatorname{null}T'\leq \operatorname{null}S+\operatorname{null}T.$

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    This looks much nicer if you draw an exact sequence, but OP is probably not familiar with that yet. – 2017-01-06
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    $\mbox{ker} T'=\mbox{ker} T$? That doesn't seem right. For example take $T$ to be the identity and $S=0$. – 2017-01-06
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    @AlexR., well, in that case $ST = 0$, $\ker ST = V$, $T' = T$ and finally, $\ker T' = \ker T$. The key here is that $\ker T\subseteq\ker ST$, so restricting will not change the kernel. – 2017-01-07
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    You can also just think of exact sequence $0\to\ker T\to\ker ST\to\ker S$. If $\ker ST\to\ker S$ were epimorphism, we would get equality $\operatorname{null}(ST) = \operatorname{null}(S)+\operatorname{null}(T)$, but since it might not be, we get desired inequality. – 2017-01-07
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Consider the vector space spanned by the direct sum of the kernels of $S,T$. For any element $x$ in this space, we can write it as $x=s+t$ where $s\in\mbox{Ker}(S)$ and $t\in\mbox{Ker}(T)$. Then:

$$STx=STs+STt=STs+0.$$

By definition, the kernel of $ST$ is the set of vectors $y$ such that either $Ty=0$ or $S(Ty)=0$. Can you finish it from here?

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    Here's what I came up with using your idea: Take any v in $Ker(ST)$, then this v must also be inside $Ker(S) + Ker(T)$. With a subspace test, it can be shown that $Ker(ST) \leq Ker(S) + Ker(T)$ It follows that $null(ST) \leq nullS + nullT - null(Ker(S) \cap Ker(T)) $ after an application of the dimensions formula, which is the result. Is this correct? – 2017-01-06