I think this is true by the following way: let $M$ be a $n$-dimensional real manifold. Consider the diagonal embedding $\Delta:M\to M\times M$, $x\mapsto (x, x)$. Using the smooth structure of $M$, we can give a complex structure on $M\times M$, and $\Delta$ became a smooth map.
Actually, I'm not certain about the bold argument. Since $\mathbb{R}^{n}\times\mathbb{R}^{n}\simeq \mathbb{R}^{2n}$ can be naturally identified with $\mathbb{C}^{n}$, I thought that we can give a complex structure using smooth structure of $M$. Is this right?
Edit: The bold argument is wrong. As Michael said in the comment, if $M$ is not orientable, then $M\times M$ is also non-orientable, hence can't admit a complex structure. More generally, if $M_{1}, M_{2}$ are non-orientable, then $M_{1}\times M_{2}$ is also non-orientable. This follows from the Kunneth theorem and top cohomology. Hence, for example, $\mathbb{RP}^{2}\times \mathbb{RP}^{2}$ can't admit a complex structure.
No, this does not work. On any pair of coordinate patches $U$ and $V$ on $M$, you can put a complex structure on $U\times V$ by identifying $U$ and $V$ with open subsets of $\mathbb{R}^n$. However, there is no reason that the transition maps between two different sets of the form $U\times V$ should be holomorphic (since all we know is that they are a product of two smooth transition maps).
However, the answer to the question in your title is still yes. By the Whitney embedding theorem, we can embed any manifold $M$ into $\mathbb{R}^m$ for some $m$, which we may assume is even (if not, just increase it by $1$). This $\mathbb{R}^m$ then has a complex structure by identifying it with $\mathbb{C}^{m/2}$.