Upon considering the group of order 12 with presentation $\langle x,y|x^4=y^3=1,y^x=y^{-1} \rangle$, I became confused by the statement "all elements of $N \setminus Z$ have order $2$" in Yves Cornulier's answer and by the statement "$sn$ has order $2$" in Geoff Robinson's answer. Probably I am being dense.
In any case, here is an answer along similar lines but using character theory. For an irreducible character $\chi$ of a finite group $G$, define $\nu_2(\chi)$ to be $1$ if $\chi$ is afforded by a real representation, $0$ if $\chi$ is not real valued, and $-1$ otherwise. Let $t$ be the number of elements of order two in $G$. As shown in Chapter 4 of Martin Isaacs' ``Character Theory of Finite Groups", the sum of $\nu_2(\chi)\chi(1)$ over all irreducibles of $G$ is $1+t$.
Now assume for contradiction that $G$ is a counterexample of minimal order to the claim that every finite subgroup of $SL_2({\mathbb R})$ is cyclic. So, every proper subgroup of $G$ is cyclic. (I believe that such groups were classified in a 1903 paper of Miller and Moreno, but we will not need that.)
As a group of odd order has no irreducible character of even degree and every reducible finite subgroup of $SL_2({\mathbb C})$ is cyclic, $|G|$ is even and and $G$ acts irreducibly on ${\mathbb C}^2$. As $SL_2({\mathbb R})$ has a unique element of order two, so does $G$. Therefore, if $G$ is a $2$-group, then $G$ is quaternion of order eight, as follows from an analysis of $2$-groups with a cyclic subgroup of index two. However, the group $Q_8$ has four linear characters, each of which satisfies $\nu_2(\chi)=1$. Since $Q_8$ has one element of order two, the unique $2$-dimensional irreducible of $Q_8$ must satisfy $\nu_2(\chi)=-1$, contradicting our assumption.
We assume now that $G$ is not a $2$-group. A Sylow $2$-subgroup $S$ of $G$ is cyclic by our minimality assumption. Thus $Aut(S)$ is a $2$-group and $S$ is central in its normalizer $N_G(S)$. By Burnside's normal $p$-complement theorem, $G$ has a normal subgroup $N$ of index $|S|$. We know that $N$ is cyclic. $S$ normalizes every Sylow subgroup of $N$. Since $G$ is not cyclic, $S$ does not centralize $N$ and therefore there is some Sylow $p$-subgroup $P$ of $N$ that is not centralized by $S$. By minimality of $|G|$, $G=SP$. An analysis of $Aut(P)$ shows that $S$ does not centralize the unique subgroup of order $p$ in $P$. By minimality of $|G|$, $|P|=p$.
As noted in Geoff Robinson's answer, some element $s$ of $S$ inverts a generator of $P$. By minimality of $|G|$, $S=\langle s \rangle$. Now $s^2$ is central in $G$ and $\langle s^2,P \rangle$ is abelian and has index two in $G$. Therefore, every irreducible of $G$ has degree at most two. If $|s|=2$ then $G$ is dihedral and has $p$ elements of order two, giving a contradiction. If $|s|>4$ then $G$ cannot have a faithful real irreducible character $\chi$ of degree two, as the image of $s^2$ must be a scalar matrix in any irreducible representation of $G$. Therefore, $|S|=4$.
Since $[G,G]=P$, $G$ has four linear characters and $p-1$ irreducible characters of degree two, as the sum of the squares of the degrees of the irreducible characters is $|G|=4p$. Two of the linear characters satisfy $\nu_2(\chi)=1$ and the other two satisfy $\nu_2(\chi)=0$.
The quotient $G/\langle s^2 \rangle$ is dihedral of order $2p$ and therefore has $(p-1)/2$ characters of degree two, each of which satisfies $\nu_2(\chi)=1$ (either by direct construction or by counting involutions). Each of these characters lifts to a (nonfaithful) irreducible of $G$. Since $G$ has a unique element of order two, the remaining $(p-1)/2$ degree two irreducibles of $G$ must satisfy $\nu_2(\chi)=-1$. Therefore, $G$ has no faithful real irreducible representation and our proof is complete.