An application of Urysohn's lemma Let $X$ be a Compact Hausdorff space. Assume that the vector space of real valued continuous functions on $X$ is finite dimensional. Show that $X$ is finite.
Suppose $X$ is infinite, given any $n\in \mathbb{N}$ we get a collection of distinct elements $x_1,\dots,x_n\in X$..
For simplicity and to study first possible non trivial case we consider $n=3$...
As $X$ is Hausdoorff for each pair of distinct elements there exists disjoint open sets containing each element..
As $x_1\neq x_2$ there exists disjoint open sets $U_{1,2}$ and $V_{1,2}$ containing $x_1$ and $x_2$ respectively...
As $x_2\neq x_3$ there exists disjoint open sets $U_{2,3}$ and $V_{2,3}$ containing $x_2$ and $x_3$ respectively...
As $x_1\neq x_3$ there exists disjoint open sets $U_{1,3}$ and $V_{1,3}$ containing $x_1$ and $x_3$ respectively...
Now consider $M_1=U_{1,2}\cap U_{1,3}, M_2=V_{1,2}\cap U_{2,3}, M_3=V_{1,3}\cap V_{2,3}$ open sets that contain $x_1,x_2,x_3$ respectively..
More over, $M_1\cap M_2=U_{1,2}\cap U_{1,3}\cap V_{1,2}\cap U_{2,3}=\emptyset$. Similarly, $M_2\cap M_3=\emptyset$  and  $M_1\cap M_3=\emptyset$..
So, we have something more than definition of Hausdorffness.. 
Given $x_1,x_2,x_3$ distinct elements, we have non empty open set $M_i$ for $1\leq i\leq 3$ which are mutually disjoint..
Now comes Uryshon's lemma..
We have $x_1$ and open set containing $x_1$ i.e., $M_1$.. See that $M_1^c$ is closed.. As $\{x_1\}$ is closed and $M_1^c$ is closed set that does not contain $x_1$ by urysohn lemma there exists continuous $f_1:X\rightarrow [0,1]$ such that $f_1(x_1)=1$ and $f_1(x)=0$ for all $x\in M_1^c$ 
We have $x_2$ and open set containing $x_2$ i.e., $M_2$.. See that $M_2^c$ is closed.. As $\{x_2\}$ is closed and $M_2^c$ is closed set that does not contain $x_2$ by urysohn lemma there exists continuous $f_2:X\rightarrow [0,1]$ such that $f_1(x_2)=1$ and $f_2(x)=0$ for all $x\in M_2^c$ 
We have $x_3$ and open set containing $x_3$ i.e., $M_3$.. See that $M_3^c$ is closed.. As $\{x_3\}$ is closed and $M_3^c$ is closed set that does not contain $x_3$ by urysohn lemma there exists continuous $f_3:X\rightarrow [0,1]$ such that $f_3(x_1)=1$ and $f_3(x)=0$ for all $x\in M_3^c$ 
Suppose that $a f_1+bf_2+cf_3=0$ for some $a,b,c\in \mathbb{R}$..
We have $af_1(x_1)+bf_2(x_1)+cf_3(x_1)=0$.. As $x_1\in M_1$ and $M_1\cap M_2=\emptyset $ we have $M_1\subset M_2^c$.. As $x_1\in M_1$ we have $x_1\in M_2^c$ and so $f_2(x_1)=0$...  As $x_1\in M_1$ and $M_1\cap M_3=\emptyset $ we have $M_1\subset M_3^c$.. As $x_1\in M_1$ we have $x_1\in M_3^c$ and so $f_3(x_1)=0$... So, $a 1+ 0 +0=0$ i.e., $a=0$..
For similar reasons, we have $b=0$ and $c=0$...
So, we have $3$ continuous functions that are linearly independent..
This is true for general $n$...
Once $n$ is fixed, intersection of finite;y many open set in case of $3$ we have $2$ open sets that contain $x_i$ is clearly open..
So, for any $n$ we get a collection of $n$ continuous function that are linearly independent..
So, if $X$ is infinite this $n$ can be made arbitrary and so there is a linearly independent subset of arbitrary cardinality so.. space of real valued continuous functions on $X$  can not be finite dimensional.. Contradiction to assumption that it  is finite dimensional..
So,$X$ can not be infinite set..
Please let me know if there are any gaps.. 
Please let me know if there are better ways of doing this or better way of writing what i have written...
 A: The logic looks right. And proofs can always be cleaner. If you wait a day and look at this one again, I'll bet you can find many improvements yourself. At least, that is how it works for me for any writing that I do.
Break out into a separate lemma before the main proof the part proving that if you have $n$ separate points in a  Hausdorff space, then there are pairwise-disjoint open sets about each.
In the main proof, you only need to say "for each $k \le n$, by Urysohn's lemma, [proper names should be capitalized] there is a function $f_k\ :\ X \to [0,1]$ with $f_k(x_k) = 1$ and $f_k(M_k^c) = \{0\}$" instead repeating the same thing over and over for different indices. You might want to start with "for each $k \le n$, $M_k^c$ is closed, and as $X$ is Hausdorff, so is $\{x_k\}$, thus by Urysohn's lemma, ..." to demonstrate to your teacher that you know the requirements for Urysohn's lemma (although, you already have left the other requirement unsaid: since $X$ is compact Hausdorff, it is normal). This would be the only reason, though. These facts are obvious or well-known and therefore usually are not explicitly mentioned.
As with the above, prove that $$f_i(x_k) = \begin{cases} 1 & i = k \\ 0 & i \ne k\end{cases}$$ for generic indices instead of repeating the same argument many times. And write the combination as $$\sum_k a_kf_k$$ for the same reason. It will be easier, not harder, for someone to read over what you have here. Too many symbols squished together in one little paragraph is difficult to follow, as your brain has to constantly switch back and forth in how it interprets what it sees.
Conclude with something like "Hence the dimension of $C(X)$ is at least $n$. Or, turning it around, the number $n$ of chosen points is at most the dimension of $C(X)$. Since that dimension is finite, there is an upper limit on the number of points that can be chosen from $X$. I.e., $X$ is finite."
A: Let $f:N\to X$ be an injection. For each $n\in N$ the point $f(n)$ does not belong to the closed set $F_n=\{f(j):j<n\}$ so there is a continuous $g_n:X\to [0,1]$ such that $g_n (f(n))=1$ and $\{g(x):x\in F_n\}\subset \{0\}.$ 
For each $n\in N,$ the function $g_n$ is linearly independent from $G_n=\{g_j:j<n\}$ because if $h$ is any linear combination of members of $G_n$ then $h(f(n))=0\ne 1=g_n(f(n)),$ so $h\ne g_n.$
