The Problem: I am trying to prove the following assertion, which I believe, but am uncertain, is true:
For a function $f:I \rightarrow E$ defined on a nondegenerate interval $I$ to a Banach space $E$, if $\lim_{x \rightarrow p^+} f(x) = y$, there exists a sequence $(y_n)$ in $E$ that converges to $y$.
My thoughts: This is very similar, but not quite the same, as a function being continuous if and only if it is sequentially continuous. Here though, we don't exactly have continuity. Now, since the limit as $x$ approaches $p$ from the right exists, one can imagine contructing a sequence in the following manner: We have that for any $\varepsilon_i$ , we can find an $x_i$ such that $|f(x_i) - p|<\varepsilon_i$. So, set $y_i = f(x_i)$ and choose successively smaller epsilons and note that $y_i \rightarrow p$ as $i \rightarrow \infty$.
Now, I'm not so sure about this argument. Does it work? Is there a better way to approach this?
To provide context, as requested, I am supplying the theorem/proof where this was encountered. I have made some slight modifications due to the author's rather nonstandard notation and terminology:
Theorem: A regulated function $f:I\rightarrow E$ from a compact perfect interval $I$ to a Banach space $E$ is bounded.
Proof: If $f$ is not bounded, we find a sequence $(x_n) \in I$ such that $\|f(x_n)\| \geq n$ for all $n \in \mathbb{N}$. Because $I$ is compact, there is a subsequence $(x_{n_k})$ of $(x_n)$ and $x \in I$ such that $x_{n_k} \rightarrow x$ as $k \rightarrow \infty$. By choosing a suitable subsequence of $(x_{n_k})$, we find a sequence $(y_n)$ that converges monotonically to $x$. Since $f$ is regulated there is a $v \in E$ with $\lim f(y_n) = v$ and therefore $\lim \|f(y_n)\|=\|v\|$. Because every convergent sequence is bounded this contradicts the assumption that $\|f(x_n)\| \geq n$.