Ok, let's construct a counterexample, since I can't express my thoughts via comments.
Start by considering the semicircle function over $\Bbb{R}$, defined by
$$x \mapsto \begin{cases}1-\sqrt{1 - x^2} & \text{if } |x| \le 1 \\ \infty & \text{if } |x| > 1 \end{cases}.$$
This is an example of an lsc which fails to admit subgradients at each point in the domain, as there is no subgradient at $x = \pm 1$. We can see this visually by observing that the points $(\pm 1, 1)$ are supported only by vertical lines.
Now, we embed this function in $\Bbb{R}^2$, then extend it to a cone. Let $C$ be the cone generated by $[-1, 1] \times \{1\}$, which is defined by the inequality $y \ge |x|$. We form a function $g$ over $C$ by defining its graph to be the (non-convex) cone generated by the graph of the semicircle function over the domain $[-1, 1] \times \{1\}$. More explicitly,
$$g(x, y) = \begin{cases}y -\sqrt{y^2 - x^2} & \text{if }(x, y) \in C \\ \infty & \text{otherwise} \end{cases}.$$
This function is scalar positive-homogeneous (only degree $1$; we shall square it soon enough!).
I also claim that $g$ is lsc and convex. I think the most intuitive way to see this is to express $\operatorname{epi} g$ as the closure of the cone generated by our semicircle defined over $[-1, 1]$. In particular,
$$\operatorname{epi} g = \overline{\operatorname{cone}} S$$
where
$$S = \left\{(x, 1, z) : |x| \le 1 \text{ and } z \ge 1 - \sqrt{1 - x^2}\right\}.$$
So, suppose $(a, b, c) \in \operatorname{cone} S$. We may therefore find $x, z, \lambda$, the latter non-negative, such that $(a, b, c) = \lambda(x, 1, z)$, $|x| \le 1$, and $z \ge 1 - \sqrt{1 - x^2}$. Clearly, $\lambda = b$. If $b = 0$, then $(a, b, c) = (0, 0, 0) \in \operatorname{epi} g$, so we may assume $b > 0$.
We also have $\left|\frac{a}{b}\right| = |x| \le 1 \implies |a| \le |b| = b$, so $(a, b, c) \in C$. Given the condition that $z \ge 1 - \sqrt{1 - x^2}$, we get
$$\frac{c}{b} \ge 1 - \sqrt{1 - \left(\frac{a}{b}\right)^2} \implies c \ge b - \sqrt{b^2 - a^2} = g(a, b) \implies (a, b, c) \in \operatorname{epi} g.$$
On the other hand, if $(a, b, c) \in \operatorname{epi g}$, then $(a, b) \in C$, hence $b \ge |a|$. Suppose for the moment that $b > 0$, and consider the point
$$(x, 1, z) = \left(\frac{a}{b}, 1, \frac{c}{b}\right).$$
We see that $b(x, 1, z) = (a, b, c)$, and by reverse engineering the above calculation, we get that $(x, 1, z) \in S$. Hence $(a, b, c) \in \operatorname{cone} S$.
On the other hand, suppose $b = 0$. Note that this implies $a = 0$ too, by definition of $C$. We know that $c \ge 0$.
Fix $\varepsilon > 0$. Consider the point $(0, \varepsilon, c)$. We have
$$c \ge 0 = g(0, \varepsilon),$$
hence this point lies in $\operatorname{epi} g$, and by the above argument, $\operatorname{cone} S$. Since this point is $\varepsilon$-close to $(a, b, c)$, it follows $(a, b, c) \in \overline{\operatorname{cone}} S$.
To finally complete the proof that $g$ is lsc and convex, note that $g$ is continuous when restricted to $C$, hence its epigraph is closed, and thus is equal to the closed convex set $\overline{\operatorname{cone}} S$.
Now, we may define $f : \Bbb{R}^2 \to [0, \infty]$ by $f(x) = g(x)^2$, where $\infty^2 = \infty$. As we are applying a monotone increasing real convex function over the positive reals to $g$, the resulting function is convex. The nice continuity properties of the squaring map also ensures that $f$ is lsc. We finally have
$$f(\lambda (x, y)) = g(\lambda (x, y))^2 = \lambda^2 g(x, y)^2 = \lambda^2 f(x, y),$$
as required. We have now completely constructed our counterexample!
The only thing left to do is prove it is a counterexample. None of the points of the form $(\pm x, x)$ will have subgradients, but let's just pick on $(1, 1)$.
Suppose $(a, b) \in \partial f(1, 1)$. Consider points along the ray $(1, 1) + t(-1, 1)$ for $t \ge 0$. Note that this ray is contained in $C$, and we have, by definition of the subgradient,
$$(a, b) \cdot (-1, 1) \le \frac{f(1 - t, 1 + t) - 1}{t}.$$
Expanding,
\begin{align*}
f(1 - t, 1 + t) &= \left(1 + t - \sqrt{(1 + t)^2 - (1 - t)^2}\right)^2 \\
&= \left(1 + t - 2\sqrt{t}\right)^2 = (1 - \sqrt{t})^4.
\end{align*}
However,
$$\lim_{t \to 0^+} \frac{(1 - \sqrt{t})^4 - 1}{t} = -\infty,$$
so for sufficiently small $t$, we have
$$\frac{f(1 - t, 1 + t) - 1}{t} < (a, b) \cdot (-1, 1)$$
against assumption. That is, $\partial f(1, 1) = \emptyset$ (finally).
Best Answer
It might be easier to consider the real case $X = \mathbb R$ first. Then, if $f(x_0) = \infty$ and $f(v) < \infty$, you never have $$\mathbb R \ni (x^*, v - x_0) \le f(v) - f(x_0) = -\infty,$$ hence $\partial f(x_0) = \emptyset$.
Even if $f(x_0) \in \mathbb R$, the subdifferential might be empty, consider $f(x) = -\sqrt{1-x^2}$ for $x \in [-1,1]$ and $f(x) = +\infty$ otherwise, at the points $x = \pm 1$.