I am a beginner student in algebraic geometry, and I wanted to understand how an non-singular complex projective variety $X$ a smooth manifold when given the subspace topology coming from $\mathcal C^n$? I can see that we can cover $X$ by affine sets but are these sets charts?
I would prefer an answer which does not use any language of scheme and is understandable with minimal knowledge of algebraic geometry.
A smooth (irreducible) variety is locally cut out by polynomials $f_1,...,f_k\in \mathbb{C}[z_1,\dots, z_n]$ whose Jacobian matrix $\left(\frac{\partial f_i}{\partial z_j}\right)$ has full rank, which is $n-r$, where $r=n-k$ is the dimension of the variety. (This is the Jacobian criterion for smoothness. This can be found in chapter 1 of Hartshorne, for instance.) These polynomial functions are, in particular, holomorphic functions, and so having a full rank Jacobian allows one to apply the holomorphic analogue of the implicit function theorem to get local holomoprhic charts for your variety.
Edit: as requested, here is a fleshing out of the last statement. I will point out that the argument is essentially the same in the $C^\infty$-manifold case, replacing holomorphic with $C^\infty$.
Let $X$ be the affine variety in question. The implicit function theorem (Proposition 1.1.11 in Huybrechts' Complex Geometry text) is the following:
Theorem. Let $U\subset \mathbb C^n$ be an open subset and let $f:U\to \mathbb{C}^k$ be a holomorphic map, where $n\geq k$. Suppose $z_0\in U$ is a point such that $$ \det\left(\frac{\partial f_i}{\partial z_j}(z_0)\right)_{1\leq i,j\leq k} \neq 0. $$ Then there exist open subsets $U_1 \subset \mathbb C^{n-k}$, $U_2\subset \mathbb{C}^k$ and a holomorphic map $g: U_1 \to U_2$ such that $U_1\times U_2 \subset U$ and $f(z) = f(z_0)$ if and only if $g(z_{k+1}, \dots, z_n) = (z_1,\dots, z_k)$.
Note that the condition on the determinant I wrote in the theorem really says that the Jacobian has maximal rank, with the extra assumption that we ordered the functions $f_1, \dots, f_k$ so that the pivot columns of the Jacobian appear in the first $k$ columns.
Furthermore, to make this a bit more intuitive, the number $k$ is the codimension of $X$, and $n-k$ is the dimension of $X$. So, you can expect that $U_1$ is going to be the coordinate chart for $X$.
Here's how: define the map $U_1 \to X\cap U$ by $(z_{k+1}, \dots, z_n) \mapsto (z_{k+1},\dots, z_n,g(z_{k+1},\dots, z_n))$. This is the map $\text{id}\times g: U_1 \to U_1 \times U_2$. (I now realize Huybrechts' notation is funny: he's writing coordinates as $z=(z_{k+1},\dots, z_n, z_1, \dots, z_k)$. I will stick to this convention to avoid clashes with his conventions.)
A point $z' = (z_{k+1},\dots, z_n, z_1, \dots, z_k)$ in the image of $\text{id}\times g$ satisfies the requirement $g(z_{k+1},\dots, z_n) = (z_1,\dots,z_k)$, which implies $$f\bigg((\text{id}\times g) (z_{k+1},\dots, z_n)\bigg) = f(z_0)=0$$ by the "if" part of the theorem. Therefore $\text{id}\times g$ is well-defined (i.e. it actually maps into $X\cap U$ and not just some random spot in $U$).
Clearly $\text{id}\times g$ is holomorphic. The inverse to $\text{id}\times g$ is the projection map $\pi: X\cap U \to U_1$, given by $\pi:(z_{k+1},\dots, z_n, z_1, \dots, z_k) =( z_{k+1},\dots, z_n)$, which is also holomorphic. To verify that it is an inverse, observe that $$ \pi \circ (\text{id}\times g) (z_{k+1},\dots, z_n) = (z_{k+1},\dots, z_n) $$ holds because $\pi$ only looks at the first $n-k$ coordinates. On the other hand, writing $z' = (z_{k+1},\dots, z_n, z_1, \dots, z_k)$ again, $$ (\text{id}\times g)\circ \pi (z') = (\text{id}\times g)(z_{k+1},\dots,z_n) = \bigg(z_{k+1},\dots, z_n, g(z_{k+1},\dots, z_n)\bigg) = z' $$ which holds by the "only if" part of the theorem: $z'\in X\cap U$ means $f(z')=0$, which says $g(z_{k+1}, \dots, z_n) = (z_1,\dots, z_k)$, granting the final equality in this equation. This establishes that $\text{id}\times g: U_1 \to X\cap U$ is a biholomorphic map, so $U_1$ is a coordinate chart for $X$.