Convergence in measure means that $\mu(\{x: |f_n(x) - f(x)| \ge \epsilon\})$ tends to $0$ as $n \to \infty$. Fix $\epsilon$ and $\eta > 0$, and choose sufficiently large $M, N > 0$ such that $\mu(\{x: |f(x)|> M\}) < \frac{\eta}{3}$ and $\mu(\{x: |f_n(x)|> M\}) < \frac{\eta}{3}$ for $n \ge N$.
(We can do this because $A_m = \{x: |f(x)| > m\}$ has $A_1 \supset ... \supset A_m \supset ...$ and the intersection $\cap_mA_m$ clearly is empty, so that since $\lim_{m \to \infty} \mu (A_m) = \mu (\cap_mA_m) = 0$, at some point we must have $\mu(A_m) < \frac{\eta}{6}$. It's in this step that we have used the finite measure property, since otherwise $\lim_{m \to \infty} \mu (A_m) = \mu (\cap_mA_m)$ isn't necessarily true. As to the $f_n$'s, for $N$ sufficiently large we have $|f_n(x)| < |f(x)| + \epsilon$ if $n \ge N$, except on a set $E_N$ with $\mu(E_N) < \frac{\eta}{6}$. To get our $M$, we take $m$ so large that $\mu(A_m) < \frac{\eta}{6}$ and thus $\mu(A_m \cup E_N) < \frac{\eta}{3}$. By selecting $M = m + \epsilon$ and $n \ge N$ we obtain $\{x: |f(x)|> M\} \subset A_m$ and $\{x: |f_n(x)|> M\} \subset A_m \cup E_N$, so we have our $M$ and $N$.)
Now $\{x: |f(x)|> M\} \cup \{x: |f_n(x)|> M\}$ has measure at most $\frac{2\eta}{3}$; on the complement of this set, we have \begin{eqnarray}|f_n(x)^2 - f(x)^2| &&=&& |f_n(x) - f(x)|\cdot|f_n(x) + f(x)| \le |f_n(x) - f(x)| \cdot \Big( |f_n(x)| + |f(x| \Big)\\ &&\le && |f_n(x) - f(x)|\cdot2M&&
\end{eqnarray}
and so by choosing $n$ large enough (and larger than $N$), we can make this expression less than $\epsilon$ on all of our remaining set, except for a part with measure less than $\frac{\eta}{3}$. Therefore $$\mu(\{x: |f_n(x)^2 - f(x)^2| \ge \epsilon\}) < \frac{2\eta}{3} + \frac{\eta}{3} = \eta$$ for $n$ large enough, where $\epsilon$ and $\eta$ were arbitrary - which proves the claim.