= Solution
The <Bolzano-Weierstrass theorem> states that every <bounded sequence> of real numbers has a <convergent subsequence>. We first use it to show that a <continuous function> on $[a,b]$ is bounded above. Otherwise, choose $x_n\in[a,b]$ with $f(x_n)>n$. A convergent subsequence $x_{n_k}\to c$ exists, with $c\in[a,b]$ because the interval is closed. <Continuity> would give $f(x_{n_k})\to f(c)$, contradicting $f(x_{n_k})>n_k\to\infty$.
Let $L=\sup\{f(x):x\in[a,b]\}$, which is finite. By the definition of <supremum>, there are $y_n\in[a,b]$ with $L-1/n<f(y_n)\le L$. Another application of the <Bolzano-Weierstrass theorem> gives a subsequence $y_{n_k}\to c\in[a,b]$. By <continuity>,
$$
f(c)=\lim_{k\to\infty}f(y_{n_k})=L.
$$
Thus \b[the supremum is actually attained]:
$$
\boxed{\exists c\in[a,b]\ \forall x\in[a,b]:\ f(x)\le f(c).}
$$
This proves the maximum part of the <extreme value theorem> from sequential compactness rather than assuming it.
Now suppose $f''<0$ on $\mathbb R$. Applying the <mean value theorem> to $f'$ shows that $f'$ is strictly decreasing. At a <local maximum> of a <differentiable function>, the left difference quotients are nonnegative and the right difference quotients are nonpositive; existence of the derivative forces $f'=0$. A strictly decreasing function has at most one zero. Therefore
$$
\boxed{f''<0\ \Longrightarrow\ \text{at most one local maximum}.}
$$
If that zero exists, $f'$ is positive to its left and negative to its right, so the local maximum is also a <global maximum>.
For the uniform bound by $K<0$, define
$$
h(x)=f'(x)-Kx.
$$
Its derivative is $h'(x)=f''(x)-K<0$, so $h$ is strictly decreasing by the <mean value theorem>. Comparing with $h(0)=f'(0)$ gives
$$
\begin{cases}
f'(x)<f'(0)+Kx,&x>0,\\
f'(x)>f'(0)+Kx,&x<0.
\end{cases}
$$
Because $K<0$, this makes $f'$ negative at sufficiently large positive $x$ and positive at sufficiently large negative $x$. The function $f'$ is continuous, since it is differentiable; the <intermediate value theorem> supplies a zero $c$ between such points. Its strict decrease makes this zero unique. Again using the <mean value theorem> for $f$, the derivative signs imply that $f$ increases up to $c$ and decreases after $c$:
$$
\boxed{f''(x)<K<0\ \text{for all }x
\ \Longrightarrow\ f\text{ has a unique global maximum}.}
$$
This illustrates <uniform negative curvature and global maximization>.
Merely requiring $f''<0$ does not force existence. Take
$$
f(x)=-e^{-x}.
$$
Then $f'(x)=e^{-x}>0$ and $f''(x)=-e^{-x}<0$, but $f$ is strictly increasing, approaches $0$ as $x\to+\infty$, and never equals $0$. \b[The answer to the final question is no], even for a function bounded above:
$$
\boxed{\sup_{\mathbb R}(-e^{-x})=0\quad\text{and the supremum is not attained}.}
$$
The stronger uniform curvature assumption prevents this behaviour. Here $f''(x)\to0$ at the end where the supremum is approached, so no uniform negative bound of that kind is available.
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