Give an example of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ that is differentiable only at $x = 7$ and continuous nowhere else. Verify both properties.
Let $$f(x) = \begin{cases} 14x-49, & \text{if $x$ is rational} \\ x^2, & \text{if $x$ is irrational} \end{cases}$$
Let $x_n = 7 - \frac{1}{n}$ and $y_n = 7 - \frac{\sqrt 2}{n}$. So, $$\lim_{n \to \infty} \frac{f(x_n) - f(7)}{x_n - 7} = \lim_{n \to \infty} \frac{14(7-\frac{1}{n}) - 49}{7-\frac{1}{n}-7} = 14$$ Similarly, $$\lim_{n \to \infty} \frac{f(y_n) - f(7)}{y_n - 7} = \lim_{n \to \infty} \frac{14(7-\frac{\sqrt 2}{n}) - 49}{7-\frac{\sqrt 2}{n}-7} = 14$$ So, the limit at $x=7$ exists, therefore $f(x)$ is continuous and differentiable at $x=7$.
Now to prove that the function is continuous nowhere else, let c be a rational number (a limit point of the set of irrationals) and let $x_n = c - \frac{\sqrt 2}{n}$. Note that $\lim_{n \to \infty} x_n = c. $ Then, $$\lim_{n \to \infty} f(x_n) = (c-\frac{\sqrt 2}{n})^2 = c^2 $$ But, $f(c) = 14(c) - 49$. These two are only equal when $c = 7$. Thus by the sequential criterion of discontinuity, $f(x)$ is discontinuous everywhere else.
Have I done this proof correctly?
No. you have proved that if $f$ is continuous at $c$ and $c$ is rational then $c=7$. You also have to show that if $f$ is continuous at $c$ and $c$ is irrational then $c=7$ (but the proof is similar to the case of rational $c$).