A uniformly continuous on the unit sphere of $c_0$ is bounded Let $S_{c_0}= \{ x\in c_0 \;: \Vert x \Vert =\;1\; \}$, where $c_0$ is the space of all sequences converging to zero and 
$$f: S_{c_0} \rightarrow \mathbb R$$
is a uniformly continuous function. Prove that $f$ is  bounded. 
I thought that if I could show $S_{c_0}$ is a compact set then by using the continuity of $f$, I would result in the requested point, but I feel like I'm missing something. 
Could anyone please help or give some hints? Thanks in advance!

You need to use that $f$ is uniform continuous. By definition, it means that: For all $\epsilon >0$, there is $\delta >0$ so that $|f(y) - f(x)| <\epsilon$ whenever $\| x- y\| <\delta$. Now pick $\epsilon =1$ and $\delta_1$ be the corresponding $\delta$ in the statement. 
Let $x\in S_{c_0}$. Let 
$$ e_1 = (1, 0, 0, \cdots ) \in S_{c_0}.$$
We will find $x = x_0, \cdots, x_N=e_1$, where $x_i \in S_{c_0}$ so that $\|x_i - x_{i-1}\| <\delta_1$ and $N$ is independent of $x$. Then
$$\begin{split}
|f(x)| &= | f(x) - f(e_1) + f(e_1)| \\ 
&= |f(x_0) - f(x_N) + f(e_1)|\\
&\le |f(x_0) - f(x_N)| + |f(e_1)| \\
&\le  |f(x_0) - f(x_{1})| + |f(x_{1}) - f(x_{2})| + \cdots + |f(x_{N-1}) - f(x_N)| + |f(e_1)| \\
&\le N + |f(e_1)|
\end{split}$$
Since $N + |f(e_1)|$ is independent of $x$, we have that $f$ is bounded. 
To find $x_1, \cdots x_N$ first we write $x = (x^1, x^2, \cdots x^n, \cdots )$ and let 
$$\begin{split} x_1 &= x + \frac{1-x^1}{N_1} e_1,\\
 x_2 &= x+\frac{2(1-x^1)}{N_1} e_1, \cdots , \\
x_{N_1} &= x+ \frac{N_1(1-x^1)}{N_1}e_1 = x+ (1-x^1)e_1. 
\end{split}$$
where $N_1$ is fixed so that $\frac{2}{N_1}<\delta_1$. Note that 
$$x_{N_1} = (1, x^2, x^3, \cdots ).$$
Next we connect $x_{N_1}$ to $e_1$. Let 
$$\begin{split}
x_{N_1 +1} &= (1,  \frac{N_1-1}{N_1} x^2, \frac{N_1 -1}{N_1} x^3 , \cdots ) \\
x_{N_1 +2} &= (1,  \frac{N_1-2}{N_1} x^2, \frac{N_1 -2}{N_1} x^3 , \cdots ) \\
&\cdots \\
x_{2N_1-1} &= (1,  \frac{1}{N_1} x^2, \frac{1}{N_1} x^3 , \cdots ) \\
x_{2N_1} &= (1, 0, 0, \cdots ) = e_1.
\end{split}$$
Thus we have found the sequence $x_1, \cdots x_N$ with $N = 2N_1$. 
(As spotted in the comment, some extra works has to be done when $x^1 = -1$. In this case one might connect it to $-e_1$, and the bound would be 
$$N + \max\{ |f(e_1)|, |f(-e_1)|\}$$
instead)
