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Let $X$ be a complete normed space, $T(t):t\ge 0$ a family of bounded operators on $X$, then $T(t)T(s)=T(t+s)$, $T(0)=I$ and $T(t)x \to x$ as $t\to 0+$ implies $t\mapsto T(t)x$ is continuous for each $x\in X$.

The assumption directly says that $T(t)x$ is continuous at $t=0$. Also, for any $t\in [0,\infty)$, $T(t+)x=\lim_{s\downarrow 0} T(t+s)x=\lim_{s\downarrow 0}T(t)T(s)x = T(t)x$ so the right limit exists and equals $T(t)x$.

But I don't see a way to argue left limit. The main problem I find is that $T(t)$ is not defined for $t<0$, so we can't really do same thing as above. If $s0$. What we want is that if $t-s$ is arbitrarily small, so is $\|T(t)x-T(s)x\|\le \|T(t)-T(s)\| \|x\|$. But how does one bound $T(t)-T(s)$?

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    Use boundedness. Suppose $\| T(t) \| \le C$ for every $t$. Let $s = t - \delta$ for some $0<\delta $\| (T(t) - T(s))x\| = \|T(t-\delta) (T(\delta) - I) x\| \le \|T(t-\delta)\| \|(T(\delta) - I)x\| \le C \|(T(\delta) - I)x\|$. As $\delta \to 0$, this goes to $0$, so $T(t^-) = T(t)$. – 2017-02-07
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    @stochasticboy321 Oh, so I need uniform boundedness for that, right? (that's where completeness of $X$ kicks in I suppose). Thanks for the hint – 2017-02-07
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    In general you use the fact that the operator norm $\|T_t\|$ can have at most exponential growth in the $t$ variable, that is, $\|T_t\|\leq Me^{\omega t}$. – 2017-02-07
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    @ByronSchmuland I thought that property requires strong continuity to prove, which I'm trying to prove here, or is it not like that? – 2017-02-07
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    @user160738 Your semigroup **is** strongly continuous. The conditions that you have on your semigroup are exactly what *strong continuity* means. Essentially, continuity on the right at zero for each $x$. – 2017-02-07
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    @ByronSchmuland I was referring to the condition that I tried to prove here as "strong continuity", but never mind that, I think I figured out the proof; I thought to get uniform boundedness near neighborhood I needed the condition I'm trying to prove here, but now I see that that's not necessary. The source of confusion was that my source defines $C_0$-semigroup in rather unconventional way (which looks like a stronger condition at first), then remarks that this is implied by (and in fact equivalent to) conventional $C_0$ semigroup definition – 2017-02-07

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Assume the space is complete so that you can apply the Uniform Boundedness Principle. To prove boundedness of $\|T(t)\|$ in a neighborhood of $0$, assume the contrary. Then for each $n=1,2,3,\cdots$, there exists $t_n$ such that $\|T(t_n)\|\ge n$ and $t_n \downarrow 0$. By the Uniform Boundedness Principle, there exists $x \in X$ such that $\|T(t_n)x\|$ is unbounded, contrary to the assumption of right continuity of $T(t)x$ at $t = 0$. Hence $\|T(t)\|$ is bounded by some $M$ in some neighborhood of $0$, say $0 \le t \lt \delta$. From the exponential property, $\|T(t)\|\le M^{n}$ for $0 \le t < n\delta$.

Therefore, for fixed $t > 0$ and $0 \le s < t$ there is a constant $C$ such that \begin{align} \|T(s)x-T(t)x\|&=\|T(s)(x-T(t-s)x)\| \\ &\le C\|x-T(t-s)x\|\rightarrow 0 \mbox{ as } s\uparrow t. \end{align}

Reference: Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations