The idea of Anton Petrunin can be made into an accurate proof. One does not need $C^1$-closeness C^2$ convergence, $C^1$ convergence is enough. Note That is, I claim that there is no $C^1$ immersion sufficiently $C^1$-close to the composition $\phi:S^2\xrightarrow{z^2}S^2\subset\Bbb R^3$.
(By the way, any map $S^2\to\Bbb R^3$ is $C^0$-close to a smooth $C^\infty$ immersionby , according to the $C^0$-dense $h$-principle (for this it is enough to know and using that $S^2$ immerses in $\Bbb R^3$).
I will denote the composition $S^2\xrightarrow{z^2}S^2\subset\Bbb R^3$ by $\phi$.
R^3$.)
It suffices to show that if $f:S^2\to\Bbb R^3$ is a self-transverse map (not necessarily an immersion) that is $C^1$-close to $\phi$ (and hence transverse to the equatorial plane), then the double point locus $\Delta_f$ of $f$ meets the equatorial plane in an odd number of points.
In order to prove this, consider the singular set $\Sigma_f$ formed by unordered pairs of distinct points of $S^2$ that get identified under $f$. This is always a manifold (as long as $f$ is self-transverse) and if $f$ has no triple points then $\Delta_f=\Sigma_f$ (but $\Delta_f$ is not a manifold in general). Now $\Sigma_f$ is a subset of the quotient $\tilde S^2=S^2\times S^2\setminus\Delta_{S^2}$ by the factor exchanging involution $T$, and it lies in a small closed tubular neighborhood $R$ of $\Sigma_\phi$ in $\tilde S^2/T$.
Note that $\Sigma_\phi$ is homeomorphic to $S^1\times\Bbb R$, with the two ends approaching two distinct points on the diagonal (the singular points of $\phi$).
The remainder of the proof can be presented in two ways.
Algebraic proof (sketch).
A simple old argument (see for instance Lemma 7 in http://arxiv.org/abs/math.GT/0305158) shows that the class of $\Sigma_f$ in $H_1^{lf}(R;\,\Bbb Z/2)$ equals the cap-product
$[\Sigma_\phi]\smallfrown w_1(T)$, where $w_1(T)$ is the first Stiefel-Whitney class of the 1-dimensional bundle associated with the double covering $\tilde S^2\to\tilde S^2/T$, and the locally-finite homology is defined in the usual way using possibly infinite chains $C_\ast^{lf}=Hom(C^\ast_c;\Bbb Z/2)$, where $C^\ast_c$ denotes cochains with compact support.
(There is a little technical problem here since $z^2:S^2\to S^2$ is not self-transverse, so to be honest one has to start with a self-transverse $C^1$-approximation of this map. Such an approximation is discussed for instance in Example 6 of the same reference.)
If you work it out you'll see that $[\Sigma_f]=[pt\times \Bbb R]\in H_1^{lf}(S^1\times\Bbb R)=H_1^{lf}(R)$, so $\Sigma_f$ cannot be a compact manifold. Hence $f$ cannot be an immersion.
Geometric proof.
It suffices to show that $\Sigma_f$ has an odd number of intersection points with the manifold $\Pi_f$ formed by unordered pairs of points of $S^2$ such that both go at least one point of the pair goes into the equatorial plane under $f$. Note that $\Pi_f\cap\Sigma_f$ lies in a small neighborhood $N$ of $\Pi_\phi\cap\Sigma_\phi$ in $\bar S^2$. This neighborhood is compact (i.e. does not approach the diagonal) since so is $\Pi_\phi\cap\Sigma_\phi$ itself (it is homeomorphic to $S^1$).
Certainly, $\#(\Sigma_{f_0}\cap\Pi_{f_0})$ is odd for some $C^1$-approximation $f_0$ of $\phi$. (You can first approximate the composition $S^1\xrightarrow{2-\text{fold covering}} S^1\subset\Bbb R^2$ by a self-transverse immersion with one double point, and then extend this approximation on the equator to both hemispheres).
If $f_1$ is another $C^1$-approximation, then it is joined to $f_0$ by a self-transverse homotopy $f_t$. The singular set of this homotopy is a bordism between $\Sigma_{f_0}$ and $\Sigma_{f_1}$. Then we also get a bordism between $\Sigma_{f_0}\cap\Pi_{f_0}$ and
$\Sigma_{f_1}\cap\Pi_{f_1}$. This bordism is a compact manifold (i.e. does not approach the diagonal) since it is contained in $N$. So $\#(\Sigma_{f_1}\cap\Pi_{f_1})$ must be odd.
But as Anton observed it cannot be odd for an immersion, so we get a contradiction.
