My question is:
Let $S = \{ (x, 0) \in \mathbb R^2: x \leq 0 \} \cup \{ (0, y) \in \mathbb R^2 : y \leq 0 \} \subset \mathbb R^2$. What is the most simple/standard/intuitive way to show that there exists no topology and smooth structure on $S$ for which the inclusion map $\iota : S \to \mathbb R^2$ is a smooth immersion?
(Of course, $\mathbb R^2$ is endowed with the standard topology and smooth structure.)
Context
I'm reviewing smooth manifold theory by working through Lee's book, having not touched any math for a long time. The question above distils what I believe to be the essential characteristics of a few exercises in the book (e.g. exercise 5.9, where you are asked to show that a square is not an immersed submanifold in $\mathbb R^2$, and exercise 5.10, where you have to show that a curve with a cusp is not an immersed submanifold in $\mathbb R^2$). Therefore, I'd be grateful if somebody could teach me a simple and standard method of tackling this problem, preferably a method that generalises to a large class of examples including the ones mentioned.
My own approach
Here is my own approach – which, as you can see, is very cumbersome.
Suppose, for contradiction, that $S$ has a topology and smooth structure for which there does exist an immersion $\iota : S \to \mathbb R^2$. Let $p = (0,0) \in S$. For every tangent vector $v \in T_p S$, there exists some smooth curve $\gamma : (-\epsilon, \epsilon) \to S$ such that $\gamma(0) = p$ and $\gamma ' (0) = v$. Then $\iota_\star(v)$, the image of $v$ under the differential $\iota_\star : T_p S\to T_{\iota(p)}(\mathbb R^2)$, is equal to $(\iota \circ \gamma)' (0)$, the tangent to the curve $\iota \circ \gamma : (-\epsilon, \epsilon) \to \mathbb R^2$ at $t = 0$.
Since the image of the curve $\iota \circ \gamma$ lies in subset $\{ (x, 0) : x \leq 0 \} \subset \mathbb R^2$, the $x$-coordinate of $\iota \circ \gamma (t)$ attains its maximum at $t = 0$. Hence the $x$-component of $(\iota \circ \gamma)' (0)$ vanishes. By a similar argument, the $y$-component of $(\iota \circ \gamma)' (0)$ vanishes too. So $\iota_\star(v) = (\iota \circ \gamma)' (0) $ is the zero vector in $T_{\iota(p)}(\mathbb R^2)$.
$\iota : S \to \mathbb R^2$ is an immersion, so $\iota_\star : T_p S \to T_{\iota(p)}(\mathbb R^2)$ is injective. But we've just shown that $\iota_\star$ maps every $v \in T_p S$ to the zero vector in $T_{\iota(p)}(\mathbb R^2)$. So $T_p S$ must be the trivial, zero-dimensional vector space. But then $S$ would have to be a manifold of dimension zero, which would imply that $S$ has the discrete topology, contradicting the requirement that $S$ needs to be second countable in order to qualify as a manifold.
That argument feels very long-winded! I'd be very grateful if someone could suggest a simpler or more standard way to tackle this problem, preferably a method that generalises to similar problems along these lines.
$\qquad$ 
Best Answer
Having slept on this problem, I have realised that my pessimism about the generalisability of my approach was unjustified. The way I went about the problem does generalise to a wide range of examples.
I thought I may as well jot down my thoughts in an answer in case somebody looks at this post in the future and has the same doubts.
For example, suppose $$ S = \{ (-t^3, t) : t \in [0, \infty) \} \cup \{ (t^3, t) : t \in [0, \infty) \} \subset \mathbb R^2$$
This is a curve with a kink at $ p = (0,0)$. We want to show that there exists no topology and smooth structure on $S$ such that the inclusion $\iota : S \to \mathbb R^2$ is a smooth immersion.
To proceed, notice that the intersection of $S$ with the unit ball $B_1(\iota(p))$ lies entirely within the wedge $\{ \mathbf x \in \mathbb R^2 : \mathbf n_1 . \mathbf x \geq 0 \} \cap \{ \mathbf x \in \mathbb R^2 : \mathbf n_2 . \mathbf x \geq 0 \}$, where $\mathbf n_1 = (\tfrac {1} {\sqrt 2}, \tfrac {1} {\sqrt 2})$ and $\mathbf n_2 = (-\tfrac {1} {\sqrt 2}, \tfrac {1} {\sqrt 2})$.
Therefore, if $\gamma : (-\epsilon, \epsilon) \to S$ is any smooth curve with $\gamma(0) = p$, then $\mathbf n_1 . \left( \iota\circ\gamma(t) \right)$ and $\mathbf n_2 . \left( \iota\circ\gamma(t) \right)$ have local minima at $t = 0$. This is enough to infer that $\mathbf n_1 . \left( (\iota\circ\gamma) ' (0) \right) = \mathbf n_2 . \left( (\iota\circ\gamma) ' (0) \right) = 0$, which implies that $(\iota\circ\gamma) ' (0) = 0$.
Since $\iota_\star : T_p S \to T_{\iota(p)}(\mathbb R^2)$ is injective, it follows that $T_p S$ is zero dimensional, leading to a contradiction as in my original post.
This approach works for just about any "curve with a kink" that I can come up with. It also works for just about any "surface with a corner" inside $\mathbb R^3$. So this approach does generalise quite well, contrary to the pessimism in my original post.