Writing a Fake Proof for Real Analysis mathematics community!
I'm teaching a course in Real Analysis soon, and one thing I wanted to include were a few "fake proofs" for my students to evaluate. The research I've done hasn't turned up any fake-proofs regarding continuity. 
So I'll put it out to you: do any of you have a favorite "proof" involving continuity or a sequential criterion with a flaw in it that could potentially stump some people?
 A: 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 $y = 1$, $x = 30/22$. 
A: 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.
