Show, if $\lim\limits_{x\longrightarrow\xi}f(x)$ exists, then $f$ is bounded in a sufficient $\mathcal{U}_{\delta}(\xi)$
Let $X$ be the domain of $f$
Hence $\lim\limits_{x\longrightarrow\xi}f(x)$ exists,
$\forall \epsilon >0 \,\,\,\exists\delta>0:\forall x,y \in \dot{\mathcal{U}}_\delta(\xi)\cap X\Longrightarrow |f(x)-f(y)|<\epsilon$
Lets now pick for any $\epsilon >0$ a $\eta\in\dot{\mathcal{U}}_\delta(\xi)$
It now holds, that $\forall \epsilon >0\,\,\,\exists\delta>0:\forall x \in \dot{\mathcal{U}}_\delta(\xi)\cap X\Longrightarrow f(x) \in \mathcal{U}_{\epsilon}(f(\eta))$
which is equivalent to:
$\forall \epsilon >0\,\,\,\exists\delta>0:\forall x \in \dot{\mathcal{U}}_\delta(\xi)\cap X\Longrightarrow f(\eta)-\epsilon<f(x)<f(\eta)+\epsilon$
So $f$ is bounded in $\dot{\mathcal{U}}_\delta(\xi)$, since $f(x) \in \mathcal{U}_{\epsilon}(f(\eta))$ and there exists
$\sup\mathcal{U}_{\epsilon}(f(\eta))=f(\eta)+\epsilon$ and $\inf\mathcal{U}_{\epsilon}(f(\eta))=f(\eta)-\epsilon$
Would be great if someone could look over it :)
I am having a little difficulty in following the proof; I think it has the right idea but it is not always clear how each step follows from you assumptions and it could be laid out more clearly; there also seems to be an assumption that $\lim f(x) = f(\eta)$, which is not necessary and may not be the case if I have misunderstood the question please forgive me, but here is another way it could be presented which I think is clearer.
Proposition Let $f$ be a function from a normed space $X$ to the normed space $Y$. If $\displaystyle \lim_{x \to \xi} f(x) $ exists then there exists an open ball of radius $\delta$ centred at $\xi$, which we denote $$ \mathcal U = \mathcal U_\delta(\xi) = \{ v \in X : \lVert v - \xi \rVert < \delta \}$$ such that $f$ is bounded on $\mathcal U$.
Proof $f(x)$ converges as $x \to \xi$. Denote the limit by $\ell \in Y$. Then for any $\varepsilon > 0$ there exists $\delta > 0 $ such that $\lVert f(x) - \ell\rVert_Y \leqslant \varepsilon$ whenever $0 < \lVert x-\xi \rVert_X \leqslant \delta$.
Choose $\varepsilon = 1$, find the corresponding $\delta$ and define $\mathcal U \subset X$ to be the open ball centred at $\xi$ of radius $\delta$ and use the triangle inequality to deduce, $$\begin{align} \lVert f(x) \rVert_Y - \lVert \ell \rVert_Y \leqslant \lVert f(x) -\ell \rVert_Y \leqslant 1 \tag 1 \end{align}$$ when $x \in \mathcal U, x \neq \xi$. Then $(1)$ can be extended to include the point $x=\xi$ and rearranged to yield the inequality, $$\lVert f(x) \rVert_Y \leqslant \max\big( 1 + \lVert \ell \rVert_Y, f(\xi) \big) $$ and it thus apparent $f$ is bounded in $\mathcal U$. Note, continuity of $f$ at $\xi$ is not required.
I hope this makes sense!