Proof of $f \in C_C(X)$ where $X$ is a metric space implies $f$ is uniformly continuous Can you tell me if the following proof is correct?
Claim:
If $f$ is a continuous and compactly supported function from a metric space $X$ into $\mathbb{R}$ then $f$ is uniformly continuous.
Proof:
The proof is in two parts.
First we want to show that $f$ is uniformly continuous on $K := \operatorname{supp}{f}$:
Let $\varepsilon > 0$.
Because $f$ is continuous we have that for each $x$ in $K$ there is a $\delta_x$ such that for all $y$ with $d(x,y) < 2 \delta_x$ we have $|f(x) - f(y)| < \varepsilon$ and because $\{ B(x, \frac{\delta_x}{2}) \}_{x \in K}$ is an open cover of $K$ there is a finite subcover which we denote $\{ B(x_i, \frac{\delta_i}{2}) \}_{i=1}^n$. 
Define $\delta := \min_i \frac{\delta_i}{2}$ and let $x$ and $y$ be any two points in $K$ with $d(x,y) < \delta$. $\{ B(x_i, \frac{\delta_i}{2}) \}_{i=1}^n$ is a cover so there exists an $i$ such that $x$ is in $B(x_i, \frac{\delta_i}{2})$ which means that $d(x,x_i) < \frac{\delta_i}{2}$. Then $d(x_i ,y) \leq d(x_i ,x) + d(x,y) < \frac{\delta_i}{2} + \delta \leq \delta_i$ hence $y$ is also in $B(x_i, \delta_i)$.
Since $d(x_i,y) < \delta_i$ and $d(x, x_i) < \delta_i$ we have $|f(x) - f(y)| \leq |f(x) - f(x_i)| + |f(x_i) - f(y)| < 2 \varepsilon$.
Next we want to show that if $f$ is uniformly continuous on $K$ then it is uniformly continuous on all of $X$:
Let $\varepsilon > 0$. For any two points $x$ and $y$ we're done if either both are in $K$ or both are outside $K$ so let $x \in X \setminus K$ and $y \in K$ with $d(x,y) < \delta$. Then there is an $i$ such that $y$ is in $B(x_i, \frac{\delta_i}{2})$. Then $d(x,x_i) \leq d(x,y) + d(y,x_i) < \delta_i$ and hence $|f(x) - f(y)| \leq |f(x) - f(x_i)| + |f(x_i) - f(y)| < 2 \varepsilon$.
Is it necessary to prove this in two parts or is the second part "obvious" and should be left away?
Thanks for your help.

That looks good except for the correction that t.b. pointed out.  In the spirit of Henning Makholm's comment, here is a "canned theorem" approach.
A continuous function on a compact metric space is uniformly continuous, so $f|_K$ is uniformly continuous.  Let $\varepsilon>0$ be given.  Then $K_\varepsilon:=\{x:|f(x)|\geq \varepsilon\}$ is a closed subset of $K$, hence compact, and $\{x:f(x)=0\}$ is a closed set disjoint from $K_\varepsilon$, so there is a positive distance $\delta_1$ between $K_\varepsilon$ and $\{x:f(x)=0\}$.  Let $\delta_2$ be such that if $x$ and $y$ are in $K$ and $d(x,y)<\delta_2$, then $|f(x)-f(y)|<\varepsilon$.  Let $\delta=\min\{\delta_1,\delta_2\}$.
If $x$ and $y$ are in $X$ and $d(x,y)<\delta$, then: 


*

*$x$ and $y$ are both in $K$, and since $d(x,y)<\delta_2$, we have $|f(x)-f(y)|<\varepsilon$; or

*one of $x$ or $y$ is not in $K$.  WLOG suppose $x$ is not in $K$.  Then $f(x)=0$, and since $d(x,y)<\delta_1$, $y$ is not in $K_\varepsilon$, meaning $|f(x)-f(y)|=|f(y)|<\varepsilon$. $\square$

