Let $S$ be a diagonalizable matrix and $S+5T=I$. Then prove that $T$ is also diagonalizable.

This proof is great!

Here's another way.

Recall that a linear transformation $F:V\to V$ is diagonalizable if and only if there exists a basis of $V$ consisting of eigenvectors of $F$.

In your case, we have two linear transformations $S,T:V\to V$ such that $$ T=\alpha(I-S) $$ where $\alpha=1/5$. Furthermore, $S$ is diagonalizable so there exists a basis $\{s_1,\dotsc,s_n\}$ of $V$ consisting of eigenvectors of $S$. If $\{\lambda_1,\dotsc,\lambda_n\}$ are the corresponding eigenvalues of $S$, then $$ Ts_k=\alpha(I-S)s_k=\alpha(s_k-Ss_k)=\alpha(s_k-\lambda_ks_k)=\alpha(1-\lambda_k)s_k $$ This proves that each $s_k$ is also an eigenvector of $T$ with corresponding eigenvalue $\alpha(1-\lambda_k)$. That is, $\{s_1,\dotsc,s_n\}$ is a basis of $V$ consisting of eigenvectors of $T$. Hence $T$ is diagonalizable.

Proposition 1. If $F$ is any field and $A\in M_{n}(F)$ is a diagonalizable matrix then, for every polynomial $f(x)$, $f(A)\in M_{n}(F)$ is also diagonalizable.

Proof. We have $A=P^{-1}DP$ for some invertible matrix $P$ and diagonal matrix $D$. Then $f(A)=P^{-1}f(D)P$. Since $f(D)$ is diagonal, $f(A)$ is diagonalizable.

Corollary. If the minimal polynomial of a matrix $A\in M_{n}(F)$ is a product of distinct linear factors then, for every polynomial $f(x)$, the minimal polynomial of $f(A)$ is also a product of distinct linear factors.

Proposition 2. If $F$ is any field and $A\in M_{n}(F)$ is a triangulable matrix then, for every polynomial $f(x)$, $f(A)\in M_{n}(F)$ is also triangulable.

Proof. We have $A=P^{-1}\Delta P$ for some invertible matrix $P$ and (upper or lower) triangular matrix $\Delta$. Then $f(A)=P^{-1}f(\Delta)P$. Since $f(\Delta)$ is (upper or lower) triangular, $f(A)$ is triangulable.

Corollary. If the minimal polynomial of a matrix $A\in M_{n}(F)$ factors completely over $F$ then, for every polynomial $f(x)$, the minimal polynomial of $f(A)$ also factors completely over $F$.