Let $R$ be a semiprime ring. Then the right socle and left socle coincide. (The right/left socle for a ring is the sum over all minimal right/left ideals of $R$.) You can assume that the left/right socle are two-sided ideals and that for a semiprime ring $R$ you have the following equivalence: Let $e=e^2\in R$ : $eR$ is a minimal right ideal iff $eRe$ is a division ring.
I didn't understand the proof in Lambeck's "lectures of rings and modules"(especially the last implication), which goes as follows:
the left socle $S$ of $R$ is $\sum Re$, where $e$ ranges over all idempotents of $R$ for which $eRe$ is a division ring. Similarly, the right socle $S'$ of $R$ is $\sum eR$. Now $S$ is an two-sided ideal, hence $S' \subseteq S$. Similarly $S \subseteq S'$.
Let's clarify what's going on: the minimal left ideals are precisely of the form $Re$. This is because of another lemma that is no doubt in your book, that a minimal left ideal $L$ satisfies $L^2=\{0\}$ or $L=Re$ for some idempotent $e$. The former case is not possible in a semiprime ring, hence we have that all minimal left ideals are of the form $Re$, and so the left socle is the sum of these $Re$'s.
You have that $eR$ is minimal iff $Re$ is minimal, owing to the symmetric nature of the condition "$eRe$ is a division ring." Let's call the set of such idempotents $X$.
Under the given assumption that $S=\sum_{e\in X}Re$ is an ideal, it must contain $eR$ for every $e\in X$, because it certainly contains every member of $X$. If $S$ contains every $eR$, then $S$ also contains $\sum_{e\in X} eR=S'$. Mutatis mutandi it also follows $S'\subseteq S$.