You are absolutely right that a complex variety with its Zariski topology is not a complex manifold, nor even a Hausdorff topological space (unless it has dimension zero).
However there is a completely canonical way of associating to a complex algebraic variety $X$ a complex analytic variety $X^{an}$.
More precisely that association Algvar $\to$ Anvar is a functor.
This functor has been studied in detail by Serre in a ground-breaking article published in 1956 and universally known by its amusing acronym GAGA.
A typical result in the article (Proposition 6, page12) is that $X$ is complete iff $X^{an}$ is compact: a highly non-trival result relying on a theorem of Chow.
In this set-up the result you are asking about can be stated as follows:
An algebraic complex variety $X$ is regular (=smooth) if and only if the associated analytic variety $X^{an}$ is a complex manifold.
Edit
Here is an English translation of GAGA.
When your base field $k$ is characteristic $0$ or algebraically closed, $X$ is always non-singular, and the morphism $X\rightarrow spec(k)$ is smooth(and etale) in these cases.
If you are in the setting of classical algebraic geometry, when $k$ is algebraically closed, the dimension of X is $0$, and the tangent space at each point is again $0$, so it is indeed non-singular.
But in modern setting, when $k$ is an arbitrary field, the answer to your question depends on whether the residue field at each $p_i$ is separable over $k$ or not. As a scheme, X is just the spectrum of a product of finite extension of your base field(assuming each $p_i$ is a closed point, but this is necessary for X to be a closed subset of $P^n$), i.e. $X= spec(\prod k_i)$, where each $k_i$ is a finite extension of $k$, your base field. When all these extensions are separable, the morphism $X\rightarrow spec(k)$ is smooth(actually, etale), and $X$ is a non-singular variety(if you call it a variety). The separable condition is always satisfied when $k$ is char $0$ or algebraically closed, but not always so in char $p$. Just like @KReiser mentioned, the case $F_p(t)/F_p(t^p)$ may happen, when you let your base field $k=F_p(t)$, and let $p$ to be the point in $P^1=Proj(k[x,y])$ corresponds to the polynomial $x^p=ty^p$.
Generally, the tangent space of a variety at a point is defined to be the dual of the stalk of the sheaf of relative differentials, i.e. $T_xX=\Omega_{X/k,x}^*$ . And smoothness is defined to be flat + relative differential locally free of the same relative dimension. When K/k separable, it is known that $Ω_{K/k}=0$.
In any case, $X$ is always a regular scheme.
Best Answer
It depends on what you mean by a variety defined over the reals.
If you mean $\operatorname{Spec} A$ where $A$ is an $\Bbb R$-algebra (or something covered by such affine opens), then the answer is yes by the standard scheme-theoretic methods - the set of singular points has at least codimension one in every irreducible component and its complement is therefore dense.
If you mean a set of the form $\{(x_1,\cdots,x_n)\in\Bbb R^n \mid f_1(x_1,\cdots,x_n) = \cdots = f_m(x_1,\cdots,x_n) = 0 \}$, the answer is no. Consider the Whitney umbrella, the surface in $\Bbb R^3$ defined by $x^2-y^2z=0$. The $z$-axis belongs to this variety, and is singular, and there are no points with $z<0$ in the umbrella but not on the $z$-axis. On the other hand, in the usual topology, no point with $z<0$ can be in the closure of a set where every point has $z\geq0$, so the negative $z$-axis is not in the closure of the smooth points of the variety in the usual topology.