So I have a partial solution to the problem below but I reached a dead end that I'm hoping someone could assist me with.
Let $V$ be a complete normed vector space.
We will use $\|\cdot\|$ to indicate the norm on $V$.
Part i) Let $(f_n)$ be a Cauchy sequence in $C([a,b],V)$. Show that for each $x \in [a,b], (f_n(x))$ is Cauchy, and so define the pointwise limit $f(x) = \lim\limits_{n \to \infty} f_n(x)$.
Solution: We know $\forall x \in [a,b], \forall \epsilon > 0, \exists N$ such that if $m,n \geq N$, then $\|f_n(x) - f_m(x)\| < \epsilon$.
So let $x \in [a,b]$. Hence, $\forall \epsilon > 0, \exists N$ such that if $n \geq N$, then $\|f_n(x) - f_m(x)\| < \epsilon$.
This shows that $(f_n(x))$ is Cauchy in $V$. Since $V$ is complete, $(f_n(x))$ is convergent. Hence, we can define $f(x) = \lim\limits_{n \to \infty} f_n(x)$.
I'm confident that I did this portion of the problem correctly, but I am not completely certain.
Part ii) Prove that $(f_n)$ converges uniformly. HINT: Use the Cauchy criterion to obtain an estimate for $\|f_n(x) - f(x)\|$ that is independent of the point $x$.
So I'm having a bit of trouble with this part.
So I know $\|f_n(x) - f(x)\| = \lim\limits_{m \to \infty}\|f_n(x) - f_m(x)\|$ but I don't know how to bound this by $\epsilon$.
Any help or corrections to the work I have done will be appreciated.
If this is word-for-word what your question says, then if I have to make an educated guess, I would say we put the sup norm on $C([a,b],V)$. The way the question is phrased does indeed suggest that the objective is to prove that $C([a,b],V)$ with sup norm is a Banach space (and usually in my experience, if nothing else is said, this is the norm we consider on the space of continuous functions).
For any function $g:[a,b]\to V$, I shall write $\lVert g \rVert_{\infty}:= \sup_{x\in [a,b]}\lVert g(x)\rVert$, where $\lVert\cdot \rVert$ is the norm on $V$. Now, by definition of $(f_n)$ being Cauchy in $C([a,b],V)$ (with sup norm), given $\epsilon>0$, there exists $N\in\Bbb{N}$ such that for all $m,n\geq N$, we have \begin{align} \lVert f_n-f_m\rVert_{\infty}\leq \epsilon. \end{align} Now, let $x\in [a,b]$ be arbitrary, and let $n\geq N$. Then, \begin{align} \lVert f_n(x)-f(x)\rVert&= \lVert f_n(x)-\lim_{m\to \infty}f_m(x)\rVert\\ &=\lim_{m\to \infty}\lVert f_n(x)-f_m(x)\rVert \tag{$*$}\\ &\leq \limsup_{m\to \infty}\lVert f_n-f_m \rVert_{\infty} \end{align} (by the way I put a $\limsup$ in the end because for a given $n$, $\lim_{m\to \infty}\lVert f_n-f_m\rVert_{\infty}$ need not exist, but the limit superior of a sequence of real numbers always exists). Try to justify why you can pull the limit out in $(*)$. Now, can you finish up the proof that $f_n\to f$ uniformly (i.e $\lVert f_n-f \rVert_{\infty}\to 0$)?