It is not hard to construct such curves explicitly, e.g. my favourite example is a curve $U$ singular at $(3:4:5i)$ and also passing through $(1:0:0)$, $(0:1:0)$, $(0:0:1)$, $(1:1:0)$, $(1:0:1)$, $(0:1:1)$. 
(Amusingly, one more real point on $U$ is also rational, namely $(3^4:4^4:5^4)$; 
two more are quadratic irrational).


Is there an easier argument demonstrating that such curves exist?
A reference would be good enough, too. 




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**Edit**: It is immediate from Bezout theorem that a curve $W$ with more than 9 real points must contain a real oval (i.e. a curve in $\mathbb{RP^2}$), just look at the intersection of $W$ and its complex conjugate $W^\*$. A similar argument (see Noam's answer) shows that if it has 8 real points it must also have 9 of them (counting multiplicities, as usual). 



projective plane cubics with exactly 9 real points
Let $P\_1,\ldots,P\_8$ be "random" real points (which could even be rational).
Then the space of cubic polynomials vanishing on all $P\_i$ has dimension
$10 - 8 = 2$. Let $(C\_1, C\_2)$ be a basis of this real vector space.
Then the cubic curves $C\_1=0$ and $C\_2=0$ meet at $P\_1,\ldots,P\_8$ 
and at some ninth point $P\_9$,
which is automatically real (and also rational if $P\_1,\ldots,P\_8$ were).
The cubic $C\_1 + i C\_2 = 0$ then passes through $P\_1,\ldots,P\_9$
and has no other real point.


