Writing a Fake Proof for Real Analysis
This does not involve continuity, but it is a "false proof" that involves basic concepts. I made this up as a problem set question when I first taught linear algebra.
What is incorrect about the following chain of reasoning?
We have this problem: $$ \left[ \begin{array}{cc} 1 & -2 \\ 3 & 2 \\ 1 & -4 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} -1 \\ 5 \\ 1 \end{array} \right] $$
Hence: $$ \left[ \begin{array}{ccc} 2 & 6 & 2 \\ -3 & 1 & 0 \end{array} \right] \left[ \begin{array}{cc} 1 & -2 \\ 3 & 2 \\ 1 & -4 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{ccc} 2 & 6 & 2 \\ -3 & 1 & 0 \end{array} \right] \left[ \begin{array}{c} -1 \\ 5 \\ 1 \end{array} \right] $$
Hence: $$ \left[ \begin{array}{cc} 22 & 0 \\ 0 & 8 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array}\right] = \left[ \begin{array}{c} 30 \\ 8 \end{array} \right] $$
Therefore, the solution is $y = 1$, $x = 30/22$.
I'm not sure I have a favorite one, but I find this one nice:
The proof of the following statement has a flaw. Identify the false statement in the proof, and give an example to show that it is false:
Every bounded continuous real-valued function $f$ on $\mathbb{R}$ attains its maximum.
Proof. Let $M=\sup\{f(x) \colon x \in \mathbb{R}\}$, and let $x^*, x_n \in \mathbb{R}$ such that $x_n \to x^*$ and $f(x_n) \to M$. Since $f$ is continuous, $f(x_n) \to f(x^*)$, which implies $f(x^*)=M$. Hence, $x^*$ is where $f$ attains its maximum.
In general, a nice place for ideas or even outright problems might be Counterexamples in Calculus.