Take for example $$
f(t) = \begin{cases}
-x^2 &\text{if $x < 0$} \\
x^2 &\text{if $x \geq 0$.}
\end{cases}
$$
For $x<0$ you have $f''(x) = -2$ while for $x > 0$ you have $f''(x) = 2$. $f$ is continuous as $0$, since $\lim_{t\to0^-} f(t) = \lim_{t\to0^+} f(t) = 0$, but since the second-order left-derivative $-2$ is different from the second-order right-derivative $2$ at zero, the second-order derivative doesn't exist there.
For your second question, maybe things are clearer if stated like this
If the second derivative is greater than zero or less than zero at some point $x$, that point cannot be an inflection point
This is quite reasonable - if the second derivative exists and is positive (negative) at some $x$, than the first derivative is continuous at $x$ and strictly increasing (decreasing) around $x$. In both cases, $x$ cannot be an inflection point, since at such a point the first derivative needs to have a local maximum or minimum.
But if the second derivative doesn't exist, then no such reasoning is possible, i.e. for such points you don't know anything about the possible behaviour of the first derivative.
There are many pieces to this question that you've asked, so let's answer them in turn.
Firstly, your reasoning about your question is very good. I'm going to reference it, so allow me to go through the argument once more.
Claim: A function $f(x)$ is continuous and twice differentiable. Prove that if the line segment joining $(a,f(a))$ and $(b,f(b))$ intersects the graph at $(c,f(c))$, then there exists a $t\in(a,b)$ such that $f''(t)=0$.
Proof: If $f$ is a line, then we are done since every point on $f$ has zero second derivative (Remark: but no inflection points).
Otherwise, the derivative of $f$ will be nonconstant. This means that the second derivative will be either or positive or negative somewhere. As you show, the second derivative will actually be both positive somewhere and negative somewhere. Since $f$ is twice-differentiable, $f'$ exists and is continuous. By the extreme value theorem, $f'$ will have a max or a min in the interval $[a,b]$. And since we know $f'$ is not always increasing/decreasing, there will be a $t \in (a,b)$ where $f'(t)$ is a max or a min (and therefore an inflection point of the graph of $f$). $\diamondsuit$
This is a slight reparsing of your proof, but it's solid. I think one good thing to clarify with your professor is what you mean by concavity. I'm going to assume that you define concave up and down along these lines:
$f$ is concave up on an interval $I$ if all tangents to the curve $f$ are below the graph of $f$ on $I$.
$f$ is concave down on an interval $I$ if all tangents to the curve $f$ are above the graph of $f$ on $I$.
Note that these definitions rely most fundamentally on a well-behaved first derivative, since tangents are properties from the first derivative. Since our first derivative exists and is continuous, we can expect nice and reasonable behavior from concavity.
If these agree with your definitions, then it's easy to see why the M-graph above cannot have no change of concavity if its first derivative is continuous. Since the first derivative exists, we can talk about tangents at all points of $x$. And since it's continuous, there can be no jumps in the slope of the tangents. In your proof, you used the mean value theorem to find a point $a' \in (a,c)$ where the slope was the same as the slope of the initial line. For ease of argument, let's suppose there is exactly one such point (i.e. I don't want to worry about many oscillations right now, but it all works), and that $f$ is above the line in $(a,c)$, for ease of description.
Now, to directly talk about the M-graph. Then if $f$ were always concave down, then the tangents to $f$ would always be above the graph of $f$. But we know that after $a'$, the slope of $f$ is always less than the slope of the $ab$-line segment. Further, we know that $f$ intersects the $ab$-line segment at $c$. Having a tangent with derivative less than the slope of the $ab$-line segment and intersecting the $ab$-line segment means that $f$ must drop below the line segment. This is a contradiction to the graph itself, and so $f$ is not always concave down.
Even in non-simplified cases, this style of argument (looking at the behavior of tangents to $f$ as they intersect and go above/below the line segment) shows that there will always be concavity changes. But behind the works we are using that they behave reasonably - if we don't know that $f'$ is continuous, then all bets are off.
Finally, $f(x) = \sin^2 x$ has many points of inflection. I'm going to guess that either your professor accidentally misspoke or had something else in mind (happens to me in front of my classes all the time).
Best Answer
As also illustrated in this answer, it is incorrect to assume that the condition $f''(x) = 0$ is the sole criteria to determine whether a point is a point of infection. I will quote the same example just to make this point.
Consider the function: $$f(x)=\cases{ -x^2 & $x\le 0$ \\ x^2 & $x>0$ }$$
If you sketch the graph of the function, you can note quite clearly that the concavity of the above function changes at $x = 0$. However, the second derivative is not continuous at that point.
Therefore, your argument that a point of inflection implies that the slope "does not change" is also quite incorrect. While it is true that if the second derivative is continuous at a point $x = a$ and the second derivative exists at the point, then $f''(a) = 0$, we cannot simply conclude a point is a point of inflexion by noting the double derivative vanishes. You can also consider the graphs of the functions $y = x^{2n}$ for $2 \le n$ where the double derivative vanishes at $x=0$ but the concavity does not change.