EDIT: I misunderstood the exercise, as pointed out by @Arthur in the comments. I only showed that a manifold does not lose orientability when a point is removed. But we can imagine a one-dimensional manifold gaining orientability when a point is removed. Here orientability would also be affected.
EDIT: In case you are looking for a solution to the exercise, here is a hint. The proof below shows (in painstaking detail!) that a manifold does not lose orientability when a point is removed. As for the second statement, we prove that an orientation on $M\setminus x$ can be extended to an orientation on $M$. There is at most one way to do this, since an orientation on a path-component is uniquely determined by its restriction to any point of the path-component (paving lemma argument). When $M$ has dimension greater than one, this extension is continuous ("locally consistent") -- use the fact $M\setminus x$ is path-connected (to see this without point-set topology, use LES to compute $H_0$), something which is not true in general when $M$ has dimension $1$.
The following is an exercise in Hatcher:
Show that deleting a point from a manifold of dimension greater than 1 does not affect orientability of the manifold.
I do not understand why the dimension must be greater than one. Is there something wrong with my proof for arbitrary dimension (below)? I have tried to be very explicit about maps, hoping to backtrace my mistake.
The heart of my argument is the commutative diagram near the bottom.
Proof. Let $z\in M$ where $M$ is an $n$-manifold. Put $M':= M\setminus\lbrace z\rbrace$. Choose an orientation $$\lambda\colon M\rightarrow M_R$$ (as in Hatcher, s. 3.3). For each $x\in M'$, excision gives a canonical isomorphism $$ \varphi_x\colon H_n(M'\vert\, x)\xrightarrow\sim H_n(M\vert \,x) $$ induced by the inclusion of pairs $(M',M'\setminus \lbrace x\rbrace )\hookrightarrow (M,M\setminus \lbrace x\rbrace )$. For each $x\in M'$, this allows us to define $$ \mu_x := \varphi_x^{-1}(\lambda_x). $$ We claim that $x\mapsto \mu_x$ defines an orientation of $M'$. Clearly $\mu_x$ generates $H_n(M'\vert\, x)$ for each $x\in M'$ (since $\varphi_x$ is an isomorphism). Hence we just need to check "local consistency", i.e. continuity of $x\mapsto \mu_x$.
For each $x\in M'$, the fact that $\lambda$ is an orientation means we can find $(B,\lambda_B)$ with $B\ni x$ an open ball of finite radius and $\lambda_B$ a generator of $H_n(M\vert\, B)$ with $$\lambda_B\vert_x = \lambda_x,$$ where I use the restriction bar "$\vert$" to indicate the map induced by the inclusion $(M,M\setminus B)\hookrightarrow (M,M\setminus\lbrace x\rbrace )$. Since any ball contained in $B$ satisfies the same consistency condition, we may assume that $B\subset M'$ and even that $\mathrm{cl}\, B\subset M'$.
(For the last assertion, I used that $M$ is Hausdorff to find open neighborhoods $U\ni x$ and $V\ni y$ with empty intersection. Then $x\in U\cap B$ and $z$ does not lie in the closure of $U\cap B$. Finally, open balls form a basis, so we can choose a smaller open ball around $x$ contained in $U\cap B$.)
The fact that $\mathrm{cl}\, B\subset M'$ means that the inclusion $(M',M'\setminus B)\hookrightarrow (M,M\setminus B)$ induces an isomorphism
$$
\psi\colon H_n(M'\vert B)\xrightarrow\sim H_n(M\vert B).
$$
Define
$$
\mu_B := \psi^{-1}(\lambda_B).
$$
We claim that $(B,\mu_B)$ is a "consistent open ball at $x$", i.e. that
$$
\mu_B\vert_y = \mu_y
$$
for each $y \in B$. But this follows from the diagram:
$$
\require{AMScd}
\begin{CD}
H_n(M'\vert\, B) @>>> H_n(M'\vert\, y) \\
@V\psi VV @VV\varphi_x V \\
H_n(M\vert\, B) @>>> H_n(M\vert\, y)
\end{CD}
$$
Here the horizontal arrows are the "restriction maps" mentioned above. All arrows in this diagram are isomorphisms (the vertical arrows are excision maps; the horizontal arrows are induced by homotopy equivalences). Furthermore, the diagram commutes since both the clockwise and the counterclockwise directions are induced by the inclusion $(M',M'\setminus B)\hookrightarrow (M,M\setminus\lbrace y\rbrace )$. But the clockwise value of $\mu_B$ is $\varphi_x (\mu_B\vert_y)$, whereas the counterclockwise value is $\psi (\mu_B)\vert_y = \lambda_B\vert_y = \lambda_y$. But
$$
\varphi_x (\mu_B\vert_y ) = \lambda_y \Rightarrow \mu_B\vert_y = \varphi_x^{-1}(\lambda_y ) = \mu_y,
$$
completing the proof. Q.E.D.
Let $M$ be a closed manifold of dimension at least 2. Let $M^*$ be the double cover of $M$ defined in Hatcher pg. 234 of the most recent edition. In cases:
i) If $M$ is orientable then $M^*$ has precisely two path components which we name $M_+$ and $M_-$ . Deleting $x$ from $M$ induces a deletion from $M^*$, but only one point say $x_+$ is deleted from $M_+$ and only one point say $x_-$ is deleted from $M_-$. Now the dimension of $M^*$ is at least two by fundamental properties of covers of manifolds, which then forces $M_+$ and $M_-$ to also be of dimension at least 2. The deletions from $M_+$ and $M_-$ simply give boundary to these path components of $M^*$, meaning $M-\{x\}$ is covered by the disjoint union of $M_+ - \{x_+\}$ and $M- \{x_-\}$, a space with precisely 2 path connected components. As this disjoint union only has 2 path components, $M-\{x\}$ is orientable
ii) this case is analogous to case (i) except we argue that deleting a point from the double cover of a non-orientable $M$ gives a cover of $M - \{x\}$ which has precisely one path component.