Prove that a linear subspace of $C([0,1])$ is closed 
Prove that the set 
  $$ W= \{f \in C([0, 1]):f(0)=0\}$$
  is a closed linear subspace of $C([0,1]).$

Here $C([0, 1])$ is the space of continuous functions equipped with the uniform norm  $||f||_\infty = \sup_{x\in[0,1]}|f(x)|$. 
I need some help in how to proceed  with this problem. 
First, I want to prove that $W$ is, indeed, a subset of $C([0,1])$. For this, we have that for any two functions $f,g\in W$, $f+g\in W$, since $f(0) + g(0) = 0$. Then, also for any $\lambda \in \mathbb{R}$, we have that $\lambda f \in W$, since $\lambda f(0) = 0$. Finally, we also have $0 \in W$. So $W$ is a linear subspace of $C([0, 1])$.
My problem is how to prove that $W$ is closed. Based on the definition of a closed set, I wanted to show that a sequence of continuous functions $f_n$ converges uniformly to $f$ and that this $f$ is also in $W$. I started with something like this:
Let $f_n$ be a sequence of continuous functions such that $f_n(0) = 0$ for any $n\in \mathbb{N}$. We need to prove that $f_n \to f$, uniformly. i.e $$ \|f_n - f\|_\infty = \sup_{x\in[0,1]} |f_n(x) - f(x)|\to 0$$ as $n\to \infty$. 
How can I proceed?

The linear functional $L(x)=x(0)$ is a continuous linear map from $C[0,1]$ to $\mathbb{C}$ because it is bounded, i.e., $|T(x)|=|x(0)| \le \|x\|$. Therefore the inverse image of $\{0\}$ under $T$ is closed, which is the subspace $W$.
