Riemann surfaces are algebraic

Solution 1:

A compact Riemann surface $X$ can always be embedded holomorphically into $\mathbb P^3( \mathbb C)$ . This is a highly non-trivial theorem.

Claim: It cannot in general be embedded into $\mathbb P^2( \mathbb C)$.
The simplest argument to support this claim is to recall that a smooth complex curve of degree $d$ in $\mathbb P^2( \mathbb C)$ has genus $g=\frac{(d-1)(d-2)}{2}$.
Since there exist Riemann surfaces of any genus $g\geq 0$ on one hand, and since on the other hand most integers are not of the form $\frac{(d-1)(d-2)}{2}$ , the claim is proved.

However any Riemann surface can be immersed, but not necessarily injectively, into $\mathbb P^2( \mathbb C)$ by skilfully (or skillfully if you prefer American English) composing an embedding into $\mathbb P^3( \mathbb C)$ with a projection onto a plane, so that the immersed curve will have nodal singularities at worst.

Bibliography
An excellent reference for these questions is Miranda's Algebraic Curves and Riemann Surfaces.
If you want to see the complete proofs of the embedding theorem of compact Riemann surfaces into $\mathbb P^3( \mathbb C)$ (which implies algebraicity of said Riemann surfaces), look at Forster's Lectures on Riemann Surfaces (Springer) or at Narasimhan's Compact Riemann Surfaces (Birkhäuser).
(Caution: the analysis used in these two books is not for the faint-hearted!)

Solution 2:

Compose the map with projections centered at points. To understand that, you can read the book Algebraic Curves and Riemann Surfaces by Rick Miranda, pages 98-102. That is a beautiful book. You will understand this perfectly.