$\{f :f\text{ is continuous and }f(0)=f(1)\}$ is Banach

Show that the subspace $S=\{f : f\text{ is continuous and }f(0)=f(1)\}$ equipped with the infinity norm is Banach.

(It's enough to show that it's closed)

My try to solve it

Take $f_n$ sequence of functions such that $f_n$ converges to $f$ in the norm Then$|f_n(x)-f(x)| \rightarrow 0$ for all $x$

Letting $x=0$

$|f_n(0)-f(0)| \rightarrow0$

$|f_n(1)-f(0)| \rightarrow0$

Can we find a finite set of bases that spans this set so we can say it's finite dimensional?

• You certainly can't say the set is finite-dimensional - for any $n$ you can have $f()$ take on arbitrary values at $i\cdot 2^{-n}$ for $1\leq i\leq 2^n-1$ and (continuously) linearly interpolate between them, so the dimension is at least $2^n-1$. Since $n$ was arbitrary... – Steven Stadnicki Nov 30 '16 at 23:22
• No, that will lead to the answer: $f(0)= \lim f_n(0) = \lim 0 = 0.$ – zhw. Nov 30 '16 at 23:23
• It is sufficient to show the complement is open. A ball in our space is {g| max{|(f-g)(x)|} is less than epsilon}. Assume f(0) is not f(1). – Jacob Wakem Dec 1 '16 at 0:08

You said it's enough to show it's closed; that's true. Note now that $f(0)=f(1)$ is equivalent to $f(0)-f(1)=0$. So your set is the preimage of $\{ 0 \}$ under $G(f)=f(0)-f(1)$. Can you conclude?
@Ian's answer already shows a nice way to finish your proof. Another approach to see $S$ is a Banach space is to realize that $S$ is isometric to $C(T)$, where $C(T)$ denotes the Banach space of continuous functions on the torus $T:=\mathbb{R}\big/\mathbb{Z}$. To prove it just define $\varphi:S\to C(T)$ by $$\varphi(f):=\tilde{f},\hspace{.5cm}\tilde{f}(\bar{x}):=f(\{x\}),$$ where $\{x\}$ denotes the fractional part of $x$. The condition $f(0)=f(1)$ in the definition of $S$ guarantees that this is well-defined. As $\varphi$ is trivially a isometry we have that $S$ is isometric to $C(T)$, and therefore a Banach space.