Let $E$ be a $\mathbb R$-Banach space, $(T(t))_{t\ge0}$ be a semigroup on $E$ and $B$ be a closed subspace of $E$ such that $$\left\|T(h)x-x\right\|_E\xrightarrow{h\to0+}0\;\;\;\text{for all }x\in B.\tag1$$ Are we able to conclude that $$\exists t>0:\sup_{s\in[0,\:t)}\left\|T(s)\right\|_{\mathfrak L(B,\:E)}<\infty?\tag2$$
My idea is as follows: Assume the contrary, i.e. $$\forall t>0:\sup_{s\in[0,\:t)}\left\|T(s)\right\|_{\mathfrak L(B,\:E)}=\infty.\tag3$$ Let $t\ge0$. By the uniform boundedness principle it holds the implication $$\left(\forall x\in B:\sup_{s\in[0,\:t)}\left\|T(s)x\right\|_E\right)<\infty\Rightarrow\sup_{s\in[0,\:t)}\left\|T(s)\right\|_{\mathfrak L(B,\:E)}<\infty\tag4.$$ So, by $(3)$, there exists a $x\in B$ such that $$\sup_{s\in[0,\:t)}\left\|T(s)x\right\|_E=\infty\tag5.$$
Are we able to conclude from $(5)$ that there is a $(h_n)_{n\in\mathbb N}\subseteq[0,t)$ with $h_n\xrightarrow{n\to\infty}0$ and $\left\|T(h_n)x\right\|_E\xrightarrow{n\to\infty}\infty$?
This would yield the desired claim, since $\left\|T(h_n)x\right\|_E\xrightarrow{n\to\infty}\left\|x\right\|$ by $(1)$ and hence we've derived a contradiction.
You're close to the right idea. The trick is to slightly upgrade your statement $(4)$ since you currently don't use the full power of $(3)$.
To use the full power of $(3)$ you don't want to fix $t \geq 0$ but rather take a whole sequence $t_n \to 0$. Indeed, if $(3)$ is true then we can find a sequence $(t_n)_{n \geq 1}$ with $0 < t_n \leq n^{-1}$ such that $\|T_{t_n}\|_{\mathcal{L}(B,E)} \geq n$. My $t_n$ will essentially fill the role of your $h_n$ except that my argument is worded a little differently for convenience.
Note however that for each $x \in B$, $(\|T_{t_n} x\|)_{n \geq 1}$ is a bounded sequence by the assumption $(1)$. Hence by the uniform boundedness theorem (applied to the restriction of the $T_{t_n}$ to $B$ so that $T_{t_n}: B \to E$), we have that $\sup_{n \geq 1} \|T_{t_n}\|_{\mathcal{L}(B,E)} < \infty$ which is a contradiction.