For a smooth closed Manifold $M^n$(of dimension $n$), if $\mathfrak{X}(M)$ denotes the set of all vector fields on the manifold $M$ then we can say that $\mathfrak{X}(M)$ form a vector space of some dimensions.
Now, let Diff$(M)$ denotes the group of all diffeomorphisms of $M$, so it is an open submanifold of $C^{\infty}(M,M)$.
Also its given in [$43. 1$ of Kriegl, Andreas; Michor, Peter W., The convenient setting of global analysis, Mathematical Surveys and Monographs. 53. Providence, RI: American Mathematical Society (AMS). x, 618 p. (1997). ZBL0889.58001.] that
The Lie Algebra of the smooth infinite dimensional Lie group Diff$(M)$ is the convenient vector space $C_c^{\infty}(M \leftarrow TM)$ of all smooth vector fields on $M$ with compact support, equipped with the negative of the usual Lie bracket.
Query:
If $\mathfrak{X}(M)$ is the lie algebra of infinite dimensional Lie group Diff$(M)$, so the lie algebra of Diff($M$)[i.e. $\mathfrak{X}(M)$] must also be infinite dimensional vector space, i.e. dim$\big(\mathfrak{X}(M)\big)=\infty$.
Now, is this statement correct or if not then please state the correct dimension of $\mathfrak{X}(M)$ and where is my thinking wrong.
(P.S. Here $\mathfrak{X}(M)=C_c^{\infty}(M \leftarrow TM)=$ the space of all smooth sections of the bundle map $p: TM \longrightarrow M$)
As long as your manifold is at least of dimension $1$, this space is infinite-dimensional. To see this, we can do the following construction:
Locally, a vector field is just a map $$ X:U \to U \times \mathbb{R}^n $$ for some open subset $U \subset \mathbb{R}^n$. Now, take a ball $B_{2\epsilon} \subset U$ of radius $2 \epsilon >0$ and consider a smooth, real valued bump function $\phi \in C^{\infty}(U)$ such that $\phi \geq 0$, $\phi=1$ on $B_{\epsilon}$ and $0$ outside of $B_{2 \epsilon}$. Now, we take any non-zero vector $z \in \mathbb{R}^n$ and define $$ X_z:U \to U \times \mathbb{R}^n, x \mapsto (x,\phi z) $$ and this extends to a smooth vector field outside of $U$ by setting it to the $0$-tangent vector.
We know that the space $C^{\infty}_c(B_{\epsilon})$ is infinite-dimensional. Hence we can construct infinitely many, linearly independent vector fields $\mathfrak{X}(M)$ by setting $$ X_{z,\varphi}:U \to U \times \mathbb{R}^n, x \mapsto (x,\varphi \phi z) $$ for $\varphi \in C^{\infty}(B_{\epsilon})$ and extending them by the $0$ vector on the compliment of $U$.