a) Let $ a,b\in\mathbb{R} $ such that $ a<b. $ According to the extreme value theorem, there is $ \zeta_1\in [a,b] $ for which $ f\left(\zeta_1\right) $ is local maximum on $ [a,b]\quad\text{(i)} . $ Consider the set $ \left\{ \delta'\in\mathbb{R}_{>0}\mid |x|>\delta'\ge \max\left( |a|,|b| \right)\implies f(x)\le f\left( \zeta_1 \right) \right\} . $ Clearly, it is bounded below by $ 0, $ and, in according to the hypothesis, it's nonvoid. So, applying Cantor's axiom, this set has an infimum $ \delta . $ Observe that $ f\left(\zeta_1\right) $ is upper bound for the restrictions $ f\mid_{\left( -\infty ,-\delta\right]} $ and $ f\mid_{\left[ \delta ,\infty\right)}\quad\text{(ii)} . $ Now, using the extreme value theorem for the intervals $ \left[ -\delta ,a\right] $ and $ \left[ b ,\delta\right] , $ we obtain $$ \zeta_2 :=\sup_{y\in\left[ -\delta ,a\right]} f(y) ,\zeta_3 :=\sup_{y\in\left[ b ,\delta\right] } f(y)\qquad\text{(iii)} . $$From $ \text{(i), (ii)} $ and $ \text{(iii)} , $ conclude that $$ \max \left( f\left( \zeta_1 \right) ,f\left( \zeta_2 \right) ,f\left( \zeta_3 \right) \right)\in\mathbb{R} $$is the global maximum. b) The previous point assures the existence of maxima. In the case that there is a maximum $ \alpha , $ and a neighbourhood $ (c,d) $ of $ \beta $ on which the function is constant and $ f\left(\beta\right) =\alpha $, let $ \epsilon_0\in \left( 0,d-c \right) $ and choose the sequences $$ x_n=\beta -\frac{\beta -c-\epsilon_0}{n} ,y_n=\beta+\frac{d-\beta -\epsilon_0}{n} ,\quad\forall n\in\mathbb{N} $$These work. Otherwise, let be a maximum at $ \zeta , $ and a sequence of real numbers $ \left( x_n\right)_{n\in\mathbb{N}} $ that converges to $ \zeta $ and $ x_n<\zeta,\forall n\in\mathbb{N} . $ The continuity of $ f $ implies that the sequence $ \left( f\left( x_n \right)\right)_{n\in\mathbb{N}} $ converges to $ f\left(\zeta\right) , $ so, according to Weierstrass, there is a divergent natural sequence $ \left( n_k \right)_{k\in\mathbb{N}} $ such that the subsequence $ \left(f\left( x_{n_k}\right) \right)_{k\in\mathbb{N}} $ is monotone. Clearly, the monotony is nondecreasing. Rewrite $ \left( x_{n_k}\right)_{k\in\mathbb{N}} $ as $ \left( x_n\right)_{n\in\mathbb{N}} . $ Now, according to the hypothesis, there is a $ \eta >\zeta $ such that $ f\left( \eta \right) <f\left( x_1 \right) . $ But, because $ f $ is continuous, it has Darboux's property. So, with this in view and given that $ f\left( x_1 \right) <f\left(\zeta\right) , $ the set $ \left\{ y\in \left( \zeta ,\eta \right)\mid f\left( y \right) =f\left( x_1 \right)\right\} $ is nonempty, has clear lower bounds, so, again, by Cantor, it admits infimum $ y_1, $ and $ y_1>\zeta . $ Furthermore, due to the monotony of $ f,\quad f\left( y_1 \right) \le f\left( x_2 \right) < f\left(\zeta\right) . $ Almost identic reasoning as above, leads us to the conclusion that the set $ \left\{ y\in \left( \zeta ,y_1 \right)\mid f\left( y \right) =f\left( x_2 \right)\right\} $ also admits an infimum $ y_2, $ and $ y_1\ge y_2>\zeta . $ Logically repeating this last procedure, we generate a nonincreasing and bounded sequence $ \left( y_n \right)_{n\in\mathbb{N}} $ so that $$ x_n<\zeta <y_n \quad\text{and}\quad f\left( y_n \right) =f\left( x_n \right) ,\qquad \forall n\in\mathbb{N} . $$By the monotone convergence theorem, $ \left( y_n \right)_{n\in\mathbb{N}} $ is convergent to a limit $ \mathcal{L} \ge\zeta . $ Suppose $ \mathcal{L} >\zeta . $ If $ f\left(\mathcal{L}\right) =f\left( \zeta \right) , $ then let be a real number $ \gamma\in\left( \zeta ,\mathcal{L} \right) $ such that $ f\left(\gamma\right) <f\left(\zeta\right) . $ Now, recall the fact that the sequence $ \left( f\left( x_n \right) \right)_{n\in\mathbb{N}} $ is convergent with base definition: $$ \forall\varepsilon\in\mathbb{R}_{>0}\quad\exists N_{\varepsilon }\in\mathbb{N}\quad n\ge N_{\varepsilon }\implies f\left(\zeta\right) -f\left( x_n \right) <\varepsilon . $$Set $ \varepsilon_0=f\left( \zeta \right) -f\left( \gamma \right) $ to get the fact that $ f\left( x_{N_{\varepsilon_0}} \right)\in\left( f\left( \gamma \right) , f\left( \zeta \right) \right) . $ But, according to the intermediate value theorem, there is a real number $ \gamma'\in\left( \zeta,\gamma \right) $ such that $ f\left( x_{N_{\varepsilon_0}} \right)=f\left( \gamma' \right) , $ which contradicts the minimality of $ y_{N_{\varepsilon_0}} . $ We're left with the result $ f\left( \mathcal{L} \right) <f\left( \zeta \right) . $ This cannot be, since $$ f\left(\mathcal{L}\right) =f\left(\lim_{n\to\infty} y_n\right) =\lim_{n\to\infty} f\left( y_n \right) = \lim_{n\to\infty} f\left( x_n \right) =f\left( \zeta \right) . $$ The assumption that $ \zeta <\mathcal{L} $ is contradictory, so, finally, the conclusion must be that $ \zeta =\mathcal{L} ,\quad\text{q.e.d.} $

Here is my solution for part a Solution We consider $f(0) > 0$, since $\lim_{x \rightarrow - \infty} f(x) = \lim_{x \rightarrow + \infty} f(x) = 0$ then there exist $a \in \mathbb{R}$ which is small enough and $b \in \mathbb{R}$ which is big enough satisfies $f(x) < f(0), \forall x < a$ and $f(x) < f(0), \forall x > b$ With $x \in [a; b]$, since $f(x)$ is continuous so there exist $c \in [a; b]$ which satisfies $f(x) \le f(c)$ and $f(c) \ge f(0)$ Combine with: $f(x) < f(0) \le f(c), \forall x (- \infty; a) \cup (b; + \infty)$ then $f(x) \le f(c), \forall x \in \mathbb{R}$ or $f(x)$ has maximum value in $\mathbb{R}$