$RP^1$ is not a regular level surface of any $C^1$ map from $RP^2$ into $R$ Since $RP^2$ is compact and connected, its continuous image in $R$ is a closed interval.  Let $f$ be this map.  Suppose $RP^1 = f^{-1}(c)$.  If $c$ is in the interior of $[a,b]$ then $RP^2$ \ $RP^1$ is not connected.  Hence $c=a$ or $c=b$.  WLOG assume $c=a$.  I next want to argue that $f$ achieving a minimum on $RP^1$ implies $df = 0$ on $RP^1$, hence not surjective.As this problem arose in the context of a differential topology problem, I am hesitant to conclude that all partial derivatives of $f$ are zero at $p$, hence $df_p$ is null.  I feel that any reference to partial derivatives requires mention of a chart.  Therefore, how, in the language of differential topology, would I rigorously phrase such a conclusion?  
Since first posting this question, I have some new thoughts.  As I said in a later comment, this is as the question appeared on a qualifying exam.  I think $f$ has to be smooth as opposed to merely $C^1$ since the Regular Level Set Theorem can only be applied with a smooth function.  If $f$ is smooth, then the tangent plane of $R$ at $c$ can be spanned by $\frac{\partial f}{ \partial x_i} \Big|_p$ and if $c$ is the minimal value $f$ achieves in $R$, then given any $v \in T_p RP^2$, curve $\gamma: [0,1] \mapsto RP^2$ such that $\gamma(0) = p$, and $\gamma'(0) = v$, then $\sum_i v_i \frac{\partial f}{ \partial x_i} \Big|_p = df_p(v) = (f \circ \gamma)'(0) = 0$, hence $\frac{\partial f}{ \partial x_i} \Big|_p = 0$ for every $i$? 

If a submanifold $N\subset M$ is the preimage of a regular value for some smooth mapping $U\to\Bbb R^k$ ($N\subset U\subset M$, $U$ open), then the normal bundle of $N$ is $M$ must be trivial. But you can easily check that the normal bundle of $\Bbb RP^1$ in $\Bbb RP^2$ is not a trivial line bundle. (You can check this either analytically or topologically.)
Let me emphasize you should not be asking about a function defined on all of $\Bbb RP^2$ to start with. :)
