The left-handed neutrino is a 2-spinor field $\eta_A$, $A=0,1$, and the Majorana mass term is a bilinear, $\Delta L = \pm 2$ term without the complex conjugation in each term,
$$ m\cdot \eta_A \eta_B \cdot \epsilon^{AB} + \text{Hermitian conjugate}$$
Note that this Majorana term is the only bilinear term without derivatives that one may construct from a single 2-component spinor (even if the complex conjugate is allowed as well).
However, in the $SU(2)$ gauge-invariant theory, $\eta_A$ is a component of a doublet, so it carries an extra index $i=1,2$ where $i=1$ is the neutrino and $i=2$ is the left-handed electron. So the Majorana term would have to be
$$ m\cdot \eta_{Aj} \eta_{Bk} \cdot \epsilon^{AB}t^{jk} + \text{Hermitian conjugate}$$
However, the only $SU(2)$-covariant tensor $t^{jk}$ with two indices of the same kind is $\epsilon^{jk}$ which is however antisymmetric, so with this choice of $t$, the term would vanish. Note that $t^{jk}=\delta^{jk}$ is not possible because $\delta$ must have one upper index and one lower index, not two indices of the same kind.
Equivalently, if there were an $SU(2)$-invariant mass term "neutrino . neutrino", by the neutrino-electron symmetry which is an element of $SU(2)$, there would also have to be a term "electron . electron" which would however violate the charge by $\Delta Q = 2$, so it clearly cannot be gauge-invariant.
One may show that the Majorana term would violate the hypercharge, too. Clearly, the neutrino left-handed field has a nonzero hypercharge $Y$, so the bilinear term $\eta\eta$ has a nonzero (double) value of the hypercharge, too, so such a term cannot be gauge-invariant – a faster way to show the same result.
