Let $f: (a, b] \to \mathbb{R}$. Prove that: $f$ is continuously resumable in $a \iff f$ is uniformly continuous.

My approach:

"$\implies$:"

Since $a$ is continuously resumable, then $a$ has to be an accumulation point and $\lim\_{x \to a} f(x)$ exists. I define another function $g$:

$$g: [a, b] \to \mathbb{R}, \left\{\begin{array}{lr}
 x \to f(x), & \text{for } x \in (a, b]\\
 \lim\_{z \to x} f(z), & \text{for } x = a\\
 \end{array}\right\}$$

$g$ is continuous on $[a, b]$ so $g$ is also uniformly continuous on $[a, b]$ and therefore also uniformly continuous on $(a, b]$. This implies $f$ is also uniformly continuous on $(a, b]$.

"$\impliedby$:"

I think this direction is quite trivial to prove. Edit: This direction is not so trivial, as Danny Pak-Keung Chan's proof shows.

My question is: Is my proof correct? Did I made a too big jump from "$g$ is uniformly continuous in $[a, b]"$ to "$g$ is uniformly continuous in $(a, b]$"?