I have a sequence of continuous functions, in the closed interval $[0,1]$, given by
$$f_n(t) = \begin{cases} 0 &\quad 0 \leq t \leq \frac{1}{2}\\ n(t-\frac{1}{2}) &\quad \frac{1}{2} < t \leq \frac{1}{2} +\frac{1}{n} \\ 1 &\quad \frac{1}{2} +\frac{1}{n} < t \leq 1 \\ \end{cases}$$
The uniform norm for two such functions $f_i(t)$, for $n = i$, and $f_j(t)$, for $n = j$ is given by
$$||f_i(t) - f_j(t)||_u = \sup_{t \in [0,1]} |f_i(t) - f_j(t)|$$
Now I would like to show that this sequence is Cauchy, using this norm specifically.
Assume $i > j$. Then in the interval $(\frac{1}{2}, \frac{1}{2} + \frac{1}{j}]$, $f_i$ will reach $1$ before $f_j$ because it has a steeper slope. $f_i$ reaches $1$ at $t = \frac{1}{2} + \frac{1}{i}$ and $f_j(\frac{1}{2} + \frac{1}{i}) = \frac{j}{i}$.
The maximum difference between the two functions then occurs at $t = \frac{1}{2} + \frac{1}{i}$ and this is the uniform norm.
$||f_i(t) - f_j(t)||_u = |1 - \frac{j}{i} | = 1 - \frac{j}{i}$.
For this to be Cauchy then it has to be case that for every $\epsilon > 0$ there has to be a sufficiently large $N$ such that for every $i, j \geq N$ we have that $||f_i(t) - f_j(t)||_u < \epsilon$.
Now this can be accomplished if N is large and $i$ and $j$ are close together in value. However if you pick a specific $j* \geq N$ and let $i$ grow without bounds then you get
$\lim_{i \to \infty} 1-\frac{j*}{i} = 1$.
So for any $N$ we can find some $i,j \geq N$ such that $i >> j$ and the uniform norm is therefore not less than the chosen $\epsilon$. Therefor you cannot use the uniform norm to prove Cauchy.
Is there anything wrong in my reasoning here?
This sequence is not Cauchy in the uniform norm. Your arguement is correct but you can also prove it without making those computations. The point-wise limit is $0$ for $x <\frac 12$ and $1$ for $x >\frac 1 2 $ making the limiting function a discontinuous function. This implies that the sequence is not Cauchy (by completeness of $C[0,1]$).