I wish to prove that all continuous functions $ [0,1] \to [0,1] $ must have a fixed point.
$ \mathcal{C}(0,1) $ is the set of all continuous functions $ [0,1] \to [0,1] $. Since $ | \mathbb{R} | = | \mathcal{C}(0,1) | $ and $ C(0,1) $ is path connected it is possible to create a continuous surjection $ U : [0,1] \to C(0,1) $.
Assume there exists a fixed point free function $ f \in \mathcal{C}(0,1) $. We can then define a function $\Delta(x) = f(U(x)(x)) $. $\Delta$ will be continuous since continuity is closed under composition.
Since $ \Delta \in \mathcal{C}(0,1) $ we then can find a $\Phi_\Delta$ such that $U(\Phi_\Delta) = \Delta$.
By substitution $ \Delta(\Phi_\Delta) = f(U(\Phi_\Delta)(\Phi_\Delta)) = f(\Delta(\Phi_\Delta)) $ which contradicts $f$ being fixpoint free hence there can not be a fix point free continuous function. $\square$
Your proof does not work, because you cannot find a continuous $U$.
There cannot be a continuous onto map from $[0,1]$ to $C(0,1)$, because $[0,1]$ is compact, and we can show that $C(0,1)$ is not compact.
We can show a space is not compact by finding a discrete infinite closed subset.
In this case, let $I_k$ be the interval $\left(\frac1{k+1},\frac 1k\right)$, and define $f_k$ so that it is zero outside $I_k$ and $1$ at the middle of $I_k$. Then the set $\{f_k\}$ is a closed infinite discrete subset of $C(0,1)$.
The quick proof (for one dimension) is more direct - it uses the intermediate value theorem.