Suppose $f:(C[0,2],\lVert\rVert_{\infty})\to (M_{2}(\mathbb{R}), \lVert T\rVert_{op})$, where $\lVert T\rVert_{op} = \sup_{\lVert \mathbf{x}\rVert =1}\lVert T\mathbf{x}\rVert$
Let $A=U^{-1}DU$ be some diagonalisable matrix, where $D$ is a diagonal matrix with entries in $[0,2]$ and define $f(g)=g(A)=U^{-1}g(D)U$. (Note: $g(D)$ means apply the function $g$ to each entry of the matrix $D$). Prove that the function $f$ is continuous.
It seems I need need to use the epsilon-delta definition, but I do not know how to simplify the expression $$\lVert f(g)-f(h)\rVert_{op}=\sup_{\lVert \mathbf{x}\rVert=1}\lVert U^{-1}[g(D)-h(D)]U\mathbf{x}\rVert$$ in order to relate it to $\lVert g-h\rVert_{\infty}$. How can I do this?
You do not tell us which norm you use on $\mathbb R^2$. But since all all norms on $\mathbb R^2$ are equivalent, we may take $\lVert \mathbf{x} \rVert = \lVert \mathbf{x} \rVert_\infty = \max(\lvert x_1 \rvert, \lvert x_2 \rvert)$.
Since the operator norm is submultiplicative (see e.g. show operator norm submultiplicative), we get
$$\lVert f(g) - f(h) \rVert_{op} = \lVert U^{-1}g(D) U - U^{-1}h(D) U \rVert_{op} = \lVert U^{-1}(g(D) - h(D)) U \rVert_{op} \\ \le \lVert U^{-1} \rVert_{op} \cdot \lVert g(D) - h(D) \rVert_{op} \cdot \lVert U \rVert_{op} = c \lVert g(D) - h(D) \rVert_{op} = c \lVert (g - h)(D) \rVert_{op} .$$ With $u = g - h$ we get $$\lVert u(D) \rVert_{op} = \sup_{\lVert \mathbf{x}\rVert=1}\lVert u(D)\mathbf{x} \rVert = \sup_{\lVert \mathbf{x}\rVert=1}\lVert \begin{pmatrix} u(d_{11})x_1\\ u(d_{22})x_2 \end{pmatrix} \rVert = \sup_{\lVert \mathbf{x}\rVert=1}\max( \lvert u(d_{11})x_1 \rvert, \lvert u(d_{22})x_2 \rvert) \\ = \max( \lvert u(d_{11}) \rvert, \lvert u(d_{22}) \rvert) \le \sup_{t \in [0,2]} \lvert u(t) \rvert = \lVert u \rVert_\infty$$ which finishes the proof.