Yes.
The keystone is:
Lemma. Let $f\colon [a,b]\to\mathbb R$ be continuous and assume that $f'_+(x)$ exists and is $>0$ for all $x\in [a,b)$. Then $f$ is strictly increasing.
Assume otherwise, i.e. $f(a)\ge f(b)$.
We recursively define a map $g\colon \operatorname{Ord}\to [a,b)$ such that $g$ and $f\circ g$ are strictly inreasing. Since the class $\operatorname{Ord}$ of ordinals is a proper class and $g$ is injective, we arrive at a contradiction, thus showing the claim.
- Let $g(0)=a$.
- For a successor $\alpha=\beta+1$ assume we have already defined $g(\beta)$. For sufficently small positive $h$ we have that $g(\beta)<g(\beta)+h<b$ and $\frac{f(g(\beta)+h)-f(g(\beta))}{h}\approx f_+(g(\beta))>0$. Pick one such $h$ and let $g(\alpha)=g(\beta)+h$.
- If $\alpha$ is a limit ordinal, assume $g(\beta)$ is defined for all $\beta<\alpha$. Let $x=\sup_{\beta<\alpha} g(\beta)$. A priori only $x\le b$, but we need $x<b$. Because $f$ is continuous and $f\circ g$ is strictly increasing, we conclude that $f(x)=\sup_{\beta<\alpha} f(g(\beta))\ge f(g(1))>f(g(0))=f(a)=f(b)$. Therefore $x<b$ as desired and we can let $g(\alpha)=x$.
$\square$
Corollary 1. (something like a one-sided Rolle theorem) Let $f\colon [a,b]\to\mathbb R$ be continuous with $f(a)=f(b)$. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=0$ for some $x\in[a,b)$.
Proof. Assume otherwise. Then either $f_+(x)>0$ for all $x$ or $f_+(x)<0$ for all $x$. In the first case the lemma applies and gives us a contradiction to $f(a)=f(b)$; in the other case, we consider $-f$ instead of $f$. $\square$
Corollary 2. (something like a one-sided IVT) Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in[a,b)$.
Proof. Apply the previous corollary to $f(x)-\frac{f(b)-f(a)}{b-a}x$. $\square$
By symmetry, we have
Corollary 3. Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_-$ exists and is continuos in $(a,b]$. Then $f'_-(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in(a,b]$. $\square$
Theorem. Let $f\in C(\mathbb R)$ be a function with $f'_-$ continuous on $\mathbb R$.
Then $f\in C^1(\mathbb R)$.
Proof.
Consider aribtrary $a\in \mathbb R$.
Let $\epsilon>0$ be given.
Then by continuity of $f'_-$, for some $\delta>0$ we have $|f'_-(x)-f'_-(a)|<\epsilon$ for all $x\in(a,a+\delta)$.
Thus for $0<h<\delta$ we have $\left|\frac{f(a+h)-f(a)}{h}-f'_-(a)\right|<\epsilon$ by corollary 3. We conclude that $f'_+(a)=f'_-(a)$, i.e. $f$ is differentiable at $a$. $\square$
Best Answer
The theorem is simply stating that the function \begin{align*} \varphi(x) &= \begin{cases} \frac{f(x) - f(a)}{x - a} & \text{if}\;x\not = a; \\ f'(a) & \text{if}\;x = a \end{cases} \end{align*} is continuous. And it clearly is; the only point to check is $x = a$, and the condition $\lim_{x\to a} \varphi(x) = \varphi(a)$ is exactly the definition of $f'(a)$. The theorem is not claiming that $f = \varphi$ everywhere on $I$.
One of the classic examples of a differentiable function $f$ with $f'$ not continuous is $f(x) = x^2\sin (1/x)$ (with $f(x) = 0$). The derivative \begin{align*} f'(x) &= \begin{cases} 2x \sin (1/x) - \cos (1/x) & \text{if}\; x \not = 0; \\ 0 & \text{if}\; x = 0 \end{cases} \end{align*} exists everywhere, but it is not continuous at $0$.